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About This Presentation
Summarize the main points, make it easy to remember and as simple as possible and don't miss out any key points and factors
Slide Content
Grade 11 Electric circuits R Toerien Page 1
Grade 11 Physics Module P5 Electric circuits
Introduction
In Grade 10 you learned about electric circuits, the relationship between the current and potential difference
in an electric circuit, and how to calculate the resistance of resistors in circuits. This year we will also learn
about the internal resistance of a battery and the rate at which work is done in an electric circuit. Below is a
summary of the basic concepts from Grade 10.
Current
Current strength (I) is measured in ampere (A) and is defined as the rate of flow of charge. One ampere
flows in a conductor when one coulomb of charge passes that point every second.
The following equation can be used to calculate current:
I =
Q
∆t
The current strength in an electric circuit is measured using an ammeter. Ammeters are connected in series
and have very low resistance so that it does not affect the total current in the circuit.
Worked example 1
1.1 How much charge flows through a conductor in 2 minutes if the current is 5 A?
Potential difference
Potential difference is the energy transferred per unit charge flowing between two points in a conductor.
One volt is equal to one joule of energy transferred per coulomb of charge.
The following equation can be used to calculate the potential difference:
V =
W
Q
Potential difference is measured using a voltmeter. Voltmeters are connected in parallel and have very high
resistance to ensure that almost no current flows through them so as not to affect the total current in the
circuit.
Name:
time (Δt)
measured in seconds (s)
charge (Q)
measured in coulomb (C)
current strength (I)
measured in ampere (A)
charge (Q)
measured in coulomb (C)
work done or energy transferred (E)
measured in joules (J)
potential difference
measured in volts (V)
Δt I
Q
time (s)
current strength (A)
charge (C)
Grade 11 Physics Module P5 Electric circuits
Introduction
In Grade 10 you learned about electric circuits, the relationship between the current and potential difference
in an electric circuit, and how to calculate the resistance of resistors in circuits. This year we will also learn
about the internal resistance of a battery and the rate at which work is done in an electric circuit. Below is a
summary of the basic concepts from Grade 10.
Current
Current strength (I) is measured in ampere (A) and is defined as the rate of flow of charge. One ampere
flows in a conductor when one coulomb of charge passes that point every second.
The following equation can be used to calculate current:
I =
Q
∆t
The current strength in an electric circuit is measured using an ammeter. Ammeters are connected in series
and have very low resistance so that it does not affect the total current in the circuit.
Worked example 1
1.1 How much charge flows through a conductor in 2 minutes if the current is 5 A?
Potential difference
Potential difference is the energy transferred per unit charge flowing between two points in a conductor.
One volt is equal to one joule of energy transferred per coulomb of charge.
The following equation can be used to calculate the potential difference:
V =
W
Q
Potential difference is measured using a voltmeter. Voltmeters are connected in parallel and have very high
resistance to ensure that almost no current flows through them so as not to affect the total current in the
circuit.
Name:
time (Δt)
measured in seconds (s)
charge (Q)
measured in coulomb (C)
current strength (I)
measured in ampere (A)
charge (Q)
measured in coulomb (C)
work done or energy transferred (E)
measured in joules (J)
potential difference
measured in volts (V)
Δt I
Q
time (s)
current strength (A)
charge (C)
Grade 11 Electric circuits R Toerien Page 2
Worked example 1 (cont.)
1.2 How much energy is transferred in a resistor if 4 C of charge flow through the resistor and a
potential difference of a 12 V is measured across the resistor.
1.3 How much work is done by a fan in 10 minutes if the current is 15 A and the potential difference is
220 V.
Resistance
Resistance is the opposition to the flow of charge and is measured in ohm (Ω). Resistance is defined as the
ratio of the potential difference across a resistor to the current in the resistor.
R =
V
I
Worked example 1 (cont.)
1.4 Calculate the resistance of a hairdryer if the current is 12 A and the potential difference is 240 V.
current strength (I)
measured in ampere (A)
potential difference
measured in volts (V)
resistance (R)
measured in ohm (Ω)
Q V
W
charge (C)
potential difference (V)
energy transferred/work done (J)
I R
V
current strength (A)
resistance (Ω)
potential difference (V)
Worked example 1 (cont.)
1.2 How much energy is transferred in a resistor if 4 C of charge flow through the resistor and a
potential difference of a 12 V is measured across the resistor.
1.3 How much work is done by a fan in 10 minutes if the current is 15 A and the potential difference is
220 V.
Resistance
Resistance is the opposition to the flow of charge and is measured in ohm (Ω). Resistance is defined as the
ratio of the potential difference across a resistor to the current in the resistor.
R =
V
I
Worked example 1 (cont.)
1.4 Calculate the resistance of a hairdryer if the current is 12 A and the potential difference is 240 V.
current strength (I)
measured in ampere (A)
potential difference
measured in volts (V)
resistance (R)
measured in ohm (Ω)
Q V
W
charge (C)
potential difference (V)
energy transferred/work done (J)
I R
V
current strength (A)
resistance (Ω)
potential difference (V)
Grade 11 Electric circuits R Toerien Page 3
Exercise 1
Investigating resistors in series and parallel
Open the following PhET simulation https://phet.colorado.edu/sims/html/circuit-construction-kit-
dc/latest/circuit-construction-kit-dc_en.html and choose Intro.
Explore the simulation. Click on the various buttons and drag different circuit components to the middle of
the page. Connect the components. Once you are confident with using the simulation, build a circuit with
the following components, and choose the electron flow option:
• two cells
• one light bulb
• one switch
1.1 Draw a circuit diagram of your circuit. Draw the switch in the closed position. (3)
1.2 Use the values from the circuit you built in QUESTION 1.1. Calculate the charge that flows through
the light bulb in one minute. You can obtain the values in your circuit by checking the ‘values’ box on
the right. (3)
1.3 Change the potential difference of the battery (by clicking on a cell and adjusting the value) and
observe what happens. Write your observations down. (2)
1.4 Choose one of the settings for potential difference in QUESTION 1.3 and calculate the energy
transferred in the light bulb in two minutes for this setting. (3)
1.5 Change the resistance of the light bulb and observe what happens. Write a sentence stating the
relationship between resistance and current and brightness of the light bulb. (2)
1.6 Figure out how to measure current using an ammeter. Measure the current in three different places
and write down what you observe about the current. Add one ammeter into your circuit diagram. (2)
Exercise 1
Investigating resistors in series and parallel
Open the following PhET simulation https://phet.colorado.edu/sims/html/circuit-construction-kit-
dc/latest/circuit-construction-kit-dc_en.html and choose Intro.
Explore the simulation. Click on the various buttons and drag different circuit components to the middle of
the page. Connect the components. Once you are confident with using the simulation, build a circuit with
the following components, and choose the electron flow option:
• two cells
• one light bulb
• one switch
1.1 Draw a circuit diagram of your circuit. Draw the switch in the closed position. (3)
1.2 Use the values from the circuit you built in QUESTION 1.1. Calculate the charge that flows through
the light bulb in one minute. You can obtain the values in your circuit by checking the ‘values’ box on
the right. (3)
1.3 Change the potential difference of the battery (by clicking on a cell and adjusting the value) and
observe what happens. Write your observations down. (2)
1.4 Choose one of the settings for potential difference in QUESTION 1.3 and calculate the energy
transferred in the light bulb in two minutes for this setting. (3)
1.5 Change the resistance of the light bulb and observe what happens. Write a sentence stating the
relationship between resistance and current and brightness of the light bulb. (2)
1.6 Figure out how to measure current using an ammeter. Measure the current in three different places
and write down what you observe about the current. Add one ammeter into your circuit diagram. (2)
Grade 11 Electric circuits R Toerien Page 4
1.7 Figure out how to measure potential difference using the voltmeter. Measure the potential difference
across the battery and across the light bulb and write down what you observe about the potential
difference. Add one voltmeter across the battery into your circuit diagram. (2)
1.8 Now add one additional light bulb in series. Ensure that the two light bulbs have the same
resistance.
• Draw a new circuit diagram. (1)
• Take the same readings for potential difference and current as above and write them in on
your circuit diagram. (2)
• Describe what happens to the current when you have more light bulbs in series. (2)
1.9 Use your original circuit and add one additional light bulb in parallel. Ensure that the two light bulbs
have different resistances.
• Draw a new circuit diagram. (1)
• Take the same readings for potential difference and current as above and write them in on
your circuit diagram. (2)
• Describe what happens to the current when you have more light bulbs in parallel. (2)
1.10 Calculate the resistance of one of the light bulbs in your parallel circuit. Verify your resistance with
the value provided by the simulation. (3)
TOTAL = 30
1.7 Figure out how to measure potential difference using the voltmeter. Measure the potential difference
across the battery and across the light bulb and write down what you observe about the potential
difference. Add one voltmeter across the battery into your circuit diagram. (2)
1.8 Now add one additional light bulb in series. Ensure that the two light bulbs have the same
resistance.
• Draw a new circuit diagram. (1)
• Take the same readings for potential difference and current as above and write them in on
your circuit diagram. (2)
• Describe what happens to the current when you have more light bulbs in series. (2)
1.9 Use your original circuit and add one additional light bulb in parallel. Ensure that the two light bulbs
have different resistances.
• Draw a new circuit diagram. (1)
• Take the same readings for potential difference and current as above and write them in on
your circuit diagram. (2)
• Describe what happens to the current when you have more light bulbs in parallel. (2)
1.10 Calculate the resistance of one of the light bulbs in your parallel circuit. Verify your resistance with
the value provided by the simulation. (3)
TOTAL = 30
Grade 11 Electric circuits R Toerien Page 5
Solving circuit problems
Resistors in series
Resistors in series are potential difference dividers. The current in a series circuit is the same everywhere.
Worked example 2
2.1 A circuit is set up according to the circuit diagram shown below. Assume that the resistance of the
battery, ammeters, and connecting wires are negligible.
Calculate the readings on the voltmeters and ammeters if VT = 12 V, R1 = 2 Ω, R2 = 1,5 Ω and R3 = 0,5 Ω.
A
2
V
1
R
1
A
1
R
2
V
2
R
3
V
3
V
T
VT = V1 + V2 + V3
I1 = I2 = I3
RT = R1
+ R2 + R3
A
2
V
1
R
1
A
3
R
2
V
2
R
3
V
3
V
T
A
1
A3
Solving circuit problems
Resistors in series
Resistors in series are potential difference dividers. The current in a series circuit is the same everywhere.
Worked example 2
2.1 A circuit is set up according to the circuit diagram shown below. Assume that the resistance of the
battery, ammeters, and connecting wires are negligible.
Calculate the readings on the voltmeters and ammeters if VT = 12 V, R1 = 2 Ω, R2 = 1,5 Ω and R3 = 0,5 Ω.
A
2
V
1
R
1
A
1
R
2
V
2
R
3
V
3
V
T
VT = V1 + V2 + V3
I1 = I2 = I3
RT = R1
+ R2 + R3
A
2
V
1
R
1
A
3
R
2
V
2
R
3
V
3
V
T
A
1
A3
Grade 11 Electric circuits R Toerien Page 6
Resistors in parallel
Resistors in parallel are current dividers. In the circuit below, the potential difference across each resistor in
parallel is the same.
Worked example 2 (cont.)
2.2 A circuit is set up according to the circuit diagram shown below. Assume that the resistance of the
battery, ammeters, and connecting wires are negligible.
Determine the readings on all the voltmeters and ammeters if VT = 12 V, R1 = 2 Ω, R2 = 1 Ω and R3 = 2 Ω.
V
3
A
1
V
1
A
T
V
T
A
3
R
2
R
3
R
1
A
2
V
2
IT = I1 + I2 + I3
VT = V1 = V2 = V3
1
R
T
=
1
R
1
+
1
R
2
+
1
R
3
V
3
A
1
V
1
A
T
V
T
A
3
R
2
R
3
R
1
A
2
V
2
Resistors in parallel
Resistors in parallel are current dividers. In the circuit below, the potential difference across each resistor in
parallel is the same.
Worked example 2 (cont.)
2.2 A circuit is set up according to the circuit diagram shown below. Assume that the resistance of the
battery, ammeters, and connecting wires are negligible.
Determine the readings on all the voltmeters and ammeters if VT = 12 V, R1 = 2 Ω, R2 = 1 Ω and R3 = 2 Ω.
V
3
A
1
V
1
A
T
V
T
A
3
R
2
R
3
R
1
A
2
V
2
IT = I1 + I2 + I3
VT = V1 = V2 = V3
1
R
T
=
1
R
1
+
1
R
2
+
1
R
3
V
3
A
1
V
1
A
T
V
T
A
3
R
2
R
3
R
1
A
2
V
2
Grade 11 Electric circuits R Toerien Page 7
Worked example 2 (cont.)
2.3 A circuit is set up according to the circuit diagram shown below. Assume that the resistance of the
battery, ammeters, switch and connecting wires are negligible.
When switch S is closed, the reading on A2 is 2 A.
2.3.1 Calculate the reading on V1, V2 and V3.
2.3.2 Calculate the reading on A1.
Exercise 2
2.1 The diagram shows an arrangement of two resistors of 12 Ω and
4 Ω. The arrangement is connected to a 6 V battery.
2.1.1 Are the resistors connected IN SERIES or PARALLEL?
2.1.2 What is the reading on the voltmeter?
2.1.3 Calculate the reading on ammeter A1 and A2.
A
1
V
3
2Ω
A
2
12Ω
6Ω
V
1
V
emf = 18 V
S
V
2
A
1
4
Ω
A
2
12
Ω
V 6 V
Worked example 2 (cont.)
2.3 A circuit is set up according to the circuit diagram shown below. Assume that the resistance of the
battery, ammeters, switch and connecting wires are negligible.
When switch S is closed, the reading on A2 is 2 A.
2.3.1 Calculate the reading on V1, V2 and V3.
2.3.2 Calculate the reading on A1.
Exercise 2
2.1 The diagram shows an arrangement of two resistors of 12 Ω and
4 Ω. The arrangement is connected to a 6 V battery.
2.1.1 Are the resistors connected IN SERIES or PARALLEL?
2.1.2 What is the reading on the voltmeter?
2.1.3 Calculate the reading on ammeter A1 and A2.
A
1
V
3
2Ω
A
2
12Ω
6Ω
V
1
V
emf = 18 V
S
V
2
A
1
4
Ω
A
2
12
Ω
V 6 V
Grade 11 Electric circuits R Toerien Page 8
The resistors are removed from the circuit and then reconnected as shown below.
2.1.4 Are the resistors connected IN SERIES or PARALLEL?
2.1.5 Calculate the new readings for each of the following:
a) A1
b) A2
c) V
2.2 A learner sets up the circuit on the right to investigate
series and parallel connections. The learner is sure
of the values of the resistors but when he looks at the
ammeters he decides that one of them must be
faulty.
2.2.1 Which ammeter is giving an incorrect reading and
what should be the correct reading be?
2.2.2 Say whether X1 and X2 are voltmeters or ammeters
and give the reading on each one. Use the corrected ammeter reading.
2.2.3 To make the cells last longer they are connected as shown in the diagram. What is the emf of each
cell used in the circuit?
2.2.4 How much current will flow through each cell when the circuit is working?
2.2.5 Why should the cells last longer when connected this way?
A
1
4 Ω
A
2
12
Ω
V 6 V
X
1
5
Ω
A
1
20
Ω
V
X
2
2 Ω
2 V
0,1A A
2
0,5A
The resistors are removed from the circuit and then reconnected as shown below.
2.1.4 Are the resistors connected IN SERIES or PARALLEL?
2.1.5 Calculate the new readings for each of the following:
a) A1
b) A2
c) V
2.2 A learner sets up the circuit on the right to investigate
series and parallel connections. The learner is sure
of the values of the resistors but when he looks at the
ammeters he decides that one of them must be
faulty.
2.2.1 Which ammeter is giving an incorrect reading and
what should be the correct reading be?
2.2.2 Say whether X1 and X2 are voltmeters or ammeters
and give the reading on each one. Use the corrected ammeter reading.
2.2.3 To make the cells last longer they are connected as shown in the diagram. What is the emf of each
cell used in the circuit?
2.2.4 How much current will flow through each cell when the circuit is working?
2.2.5 Why should the cells last longer when connected this way?
A
1
4 Ω
A
2
12
Ω
V 6 V
X
1
5
Ω
A
1
20
Ω
V
X
2
2 Ω
2 V
0,1A A
2
0,5A
Grade 11 Electric circuits R Toerien Page 9
Ohm’s Law
Ohm's law states that the potential difference across a conductor is directly proportional to the current in the
conductor. This means that when the potential difference increases, for example when more cells are
added in series in a circuit, the current strength will also increase in the same proportion. The
proportionality
V
I
is constant and is the resistance (R) of the conductor. Ohm’s Law therefore only applies
when the resistance is constant.
Factors affecting the resistance
Ohmic and non-ohmic conductors
Some devices, like batteries and diodes, do not obey Ohm’s Law and are called non-ohmic conductors.
Batteries, for example, deliver a fairly fixed voltage despite a large change in current. The filament of a light
bulb is also an example of a non-ohmic conductor. If the supply voltage to a light bulb in a circuit is
increased, the resulting increase in current causes the filament to heat up, which increases its resistance.
This effectively limits the increase in current and the V-I ratio is no longer a constant.
Most metals share this property of resistance increasing with temperature, but the change is generally
negligible and they will act as ohmic conductors. Some metals, like tungsten and nichrome, are chosen for
their ability to heat up quickly, making them effective materials to give off light or heat. If a graph of potential
difference versus current strength is drawn for a non-ohmic conductor, a curve will be observed, whereas
an ohmic conductor will form a straight-line graph.
Ohmic conductor Non-ohmic conductor
In the non-ohmic conductor, as the current increases, the gradient of the graph increases. This indicates an
increase in the resistance of the resistor since the
V
I
ratio increases.
Length
The shorter the wire, the lower the resistance.
Thickness
The thicker the resistor the lower the resistance.
Material
Different types of material have difference resistances.
Temperature
The hotter the resistor, the higher the resistance.
V
I
V
I
Ohm’s Law
Ohm's law states that the potential difference across a conductor is directly proportional to the current in the
conductor. This means that when the potential difference increases, for example when more cells are
added in series in a circuit, the current strength will also increase in the same proportion. The
proportionality
V
I
is constant and is the resistance (R) of the conductor. Ohm’s Law therefore only applies
when the resistance is constant.
Factors affecting the resistance
Ohmic and non-ohmic conductors
Some devices, like batteries and diodes, do not obey Ohm’s Law and are called non-ohmic conductors.
Batteries, for example, deliver a fairly fixed voltage despite a large change in current. The filament of a light
bulb is also an example of a non-ohmic conductor. If the supply voltage to a light bulb in a circuit is
increased, the resulting increase in current causes the filament to heat up, which increases its resistance.
This effectively limits the increase in current and the V-I ratio is no longer a constant.
Most metals share this property of resistance increasing with temperature, but the change is generally
negligible and they will act as ohmic conductors. Some metals, like tungsten and nichrome, are chosen for
their ability to heat up quickly, making them effective materials to give off light or heat. If a graph of potential
difference versus current strength is drawn for a non-ohmic conductor, a curve will be observed, whereas
an ohmic conductor will form a straight-line graph.
Ohmic conductor Non-ohmic conductor
In the non-ohmic conductor, as the current increases, the gradient of the graph increases. This indicates an
increase in the resistance of the resistor since the
V
I
ratio increases.
Length
The shorter the wire, the lower the resistance.
Thickness
The thicker the resistor the lower the resistance.
Material
Different types of material have difference resistances.
Temperature
The hotter the resistor, the higher the resistance.
V
I
V
I
Grade 11 Electric circuits R Toerien Page 10
Exercise 3 Ohm’s Law verification experiment
In this activity, you are going to investigate the relationship between current strength and potential
difference in a series circuit. Access the following PhET simulation:
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html and
choose Lab.
1. Build the following circuit using the simulation.
• One 5 V cell (click on the cell to adjust the voltage)
• One 10 Ω resistor (click on the resistor to adjust the resistance)
• A voltmeter across the battery
• An ammeter measuring the total current
• Ensure that the wire resistivity and battery resistance is as low as
possible.
2. Add one additional 5 V cell at a time in series and complete the table
below. Redraw the table in your workbook and complete the values.
Number of cells in
series
Voltmeter reading (V)
across the battery
Ammeter reading (A) V
I
1 5 0,5
2
3
4
5
3. Calculate the
V
I
values in the last column.
4. What do you notice about the
V
I
values in the
table?
5. Plot the values in the table on a graph and
draw a line of best fit. Choose the appropriate
dependent (y-axis) and independent (x-axis)
variables.
6. Determine the gradient of the line.
7. Which physical quantity is represented by the
gradient of the line?
8. This experiment forms the basis of an important law. Name and state this law.
Exercise 3 Ohm’s Law verification experiment
In this activity, you are going to investigate the relationship between current strength and potential
difference in a series circuit. Access the following PhET simulation:
https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html and
choose Lab.
1. Build the following circuit using the simulation.
• One 5 V cell (click on the cell to adjust the voltage)
• One 10 Ω resistor (click on the resistor to adjust the resistance)
• A voltmeter across the battery
• An ammeter measuring the total current
• Ensure that the wire resistivity and battery resistance is as low as
possible.
2. Add one additional 5 V cell at a time in series and complete the table
below. Redraw the table in your workbook and complete the values.
Number of cells in
series
Voltmeter reading (V)
across the battery
Ammeter reading (A) V
I
1 5 0,5
2
3
4
5
3. Calculate the
V
I
values in the last column.
4. What do you notice about the
V
I
values in the
table?
5. Plot the values in the table on a graph and
draw a line of best fit. Choose the appropriate
dependent (y-axis) and independent (x-axis)
variables.
6. Determine the gradient of the line.
7. Which physical quantity is represented by the
gradient of the line?
8. This experiment forms the basis of an important law. Name and state this law.
Grade 11 Electric circuits R Toerien Page 11
Energy and Power in a circuit
Power dissipated by a component in a circuit is the work done or energy ‘used’ per second. The energy is
not really used up but rather converted from one form to another, for example, electrical energy to light and
heat energy in the filament of a light bulb. The filament is a resistor and work is done to move the charges
through the resistor. Power (P) is the work done (W) per unit time (Δt) and is measured in watt (W). The
following equation can be used to calculate power:
Power (P) =
Energy (W)
Time (∆t)
Power
P =
W
∆t
, but W = VIΔt, and if this is substituted, we get P = VI.
V = IR and can again be substituted into P = VI to give P = I
2
R
I =
V
R
can be substituted into P = VI to give P =
V
2
R
Energy
V =
W
Q
which can be rearranged to give W = VQ
Q = IΔt and can be substituted into W = VQ to give W = VIΔt
V = IR can be substituted into W = VIΔt to give W = I
2
RΔt
I =
V
R
can be substituted into W = VIΔt to give W =
V
2
R
Δt
Worked example 4
4.1 A circuit is set up as shown in the diagram. Switch S is open.
The fuse, ammeter, battery and wires have negligible resistance.
4.1.1 Calculate the resistance of the parallel network.
4.1.2 Calculate the reading on the ammeter and the voltmeter.
4.1.3 Calculate the amount of heat released by the 12 Ω resistor
in 5 minutes.
4.1.4 Calculate the current in the 4 Ω resistor.
4.1.5 When switch S is closed, calculate the current in the circuit.
4.1.6 What will happen to the 2 A fuse when switch S is closed.
measured in seconds (s)
measured in joule (J)
measured in watt (W)
Note: The equations
for power and
energy are provided
on the data sheet so
you don’t have to
memorise them.
A
9
Ω
4
Ω
12
Ω
3
Ω
S
V
12 V
2 A
fuse
Energy and Power in a circuit
Power dissipated by a component in a circuit is the work done or energy ‘used’ per second. The energy is
not really used up but rather converted from one form to another, for example, electrical energy to light and
heat energy in the filament of a light bulb. The filament is a resistor and work is done to move the charges
through the resistor. Power (P) is the work done (W) per unit time (Δt) and is measured in watt (W). The
following equation can be used to calculate power:
Power (P) =
Energy (W)
Time (∆t)
Power
P =
W
∆t
, but W = VIΔt, and if this is substituted, we get P = VI.
V = IR and can again be substituted into P = VI to give P = I
2
R
I =
V
R
can be substituted into P = VI to give P =
V
2
R
Energy
V =
W
Q
which can be rearranged to give W = VQ
Q = IΔt and can be substituted into W = VQ to give W = VIΔt
V = IR can be substituted into W = VIΔt to give W = I
2
RΔt
I =
V
R
can be substituted into W = VIΔt to give W =
V
2
R
Δt
Worked example 4
4.1 A circuit is set up as shown in the diagram. Switch S is open.
The fuse, ammeter, battery and wires have negligible resistance.
4.1.1 Calculate the resistance of the parallel network.
4.1.2 Calculate the reading on the ammeter and the voltmeter.
4.1.3 Calculate the amount of heat released by the 12 Ω resistor
in 5 minutes.
4.1.4 Calculate the current in the 4 Ω resistor.
4.1.5 When switch S is closed, calculate the current in the circuit.
4.1.6 What will happen to the 2 A fuse when switch S is closed.
measured in seconds (s)
measured in joule (J)
measured in watt (W)
Note: The equations
for power and
energy are provided
on the data sheet so
you don’t have to
memorise them.
A
9
Ω
4
Ω
12
Ω
3
Ω
S
V
12 V
2 A
fuse
Grade 11 Electric circuits R Toerien Page 12
Paying for electricity
Household appliances need energy to operate. Each appliance has a power rating which states how much
energy is converted per second, for example, a 1500 W hairdryer will convert 1500 joules of energy every
second. Energy used in the household is calculated using a modified version of the power equation:
Energy used by an appliance = power (measured in kilowatt) x time (measured in hours)
W = PΔt
The unit for energy used in the household is therefore the kilowatt-hour (kWh) and can be calculated for
any appliance. The energy we use in a household is commonly referred to as electricity and is bought per
unit as shown in the following equation:
Cost of electricity = number of units x cost per unit
Worked example 4 (cont.)
4.2 How much does it cost to burn a 100 W light bulb for 3 hours if the cost is R2,10 per unit?
Exercise 4
4.1 Find out how many units of electricity your household uses per month.
4.1.1 Calculate the number of units used per day.
4.1.2 Assume you have 20 energy-saving light bulbs, each with a power rating of 15 W, in your
household. The 20 light bulbs are used for 4 hours every day. What percentage of your
household’s daily energy is used by the light bulbs?
Paying for electricity
Household appliances need energy to operate. Each appliance has a power rating which states how much
energy is converted per second, for example, a 1500 W hairdryer will convert 1500 joules of energy every
second. Energy used in the household is calculated using a modified version of the power equation:
Energy used by an appliance = power (measured in kilowatt) x time (measured in hours)
W = PΔt
The unit for energy used in the household is therefore the kilowatt-hour (kWh) and can be calculated for
any appliance. The energy we use in a household is commonly referred to as electricity and is bought per
unit as shown in the following equation:
Cost of electricity = number of units x cost per unit
Worked example 4 (cont.)
4.2 How much does it cost to burn a 100 W light bulb for 3 hours if the cost is R2,10 per unit?
Exercise 4
4.1 Find out how many units of electricity your household uses per month.
4.1.1 Calculate the number of units used per day.
4.1.2 Assume you have 20 energy-saving light bulbs, each with a power rating of 15 W, in your
household. The 20 light bulbs are used for 4 hours every day. What percentage of your
household’s daily energy is used by the light bulbs?
Grade 11 Electric circuits R Toerien Page 13
4.2 A toaster and a kettle are connected in a circuit as shown in the
diagram below. The toaster is rated at 800W and the kettle is rated
at 2000 W. Both are working under optimal conditions.
4.2.1 Calculate the current passing through the kettle.
4.2.2 Calculate the resistance of the kettle’s element. Show two different methods.
4.2.3 Calculate the energy consumed by the toaster in 3 minutes.
4.3 Three electrical devices, X, Y and Z, are connected to a 20 V battery with negligible internal
resistance as shown in the circuit diagram below. The power rating of
each of the devices X and Y are indicated in the diagram. Device Z is a
measuring device.
With switch S1 closed and S2 open, the devices function as rated.
4.3.1 Calculate the current in X
4.3.2 Calculate the resistance of Y
Now switch S2 is also closed.
4.3.3 Identify device Z which, when placed in the position shown, can still enable X and Y to
operate as rated. Assume that the resistances of all the devices remain unchanged.
4.3.4 Explain how you arrived at the answer to QUESTION 4.3.3.
240V
kettle
toaster
A
Z
Y
S
2
S
1
20V
X
100 W
150 W
4.2 A toaster and a kettle are connected in a circuit as shown in the
diagram below. The toaster is rated at 800W and the kettle is rated
at 2000 W. Both are working under optimal conditions.
4.2.1 Calculate the current passing through the kettle.
4.2.2 Calculate the resistance of the kettle’s element. Show two different methods.
4.2.3 Calculate the energy consumed by the toaster in 3 minutes.
4.3 Three electrical devices, X, Y and Z, are connected to a 20 V battery with negligible internal
resistance as shown in the circuit diagram below. The power rating of
each of the devices X and Y are indicated in the diagram. Device Z is a
measuring device.
With switch S1 closed and S2 open, the devices function as rated.
4.3.1 Calculate the current in X
4.3.2 Calculate the resistance of Y
Now switch S2 is also closed.
4.3.3 Identify device Z which, when placed in the position shown, can still enable X and Y to
operate as rated. Assume that the resistances of all the devices remain unchanged.
4.3.4 Explain how you arrived at the answer to QUESTION 4.3.3.
240V
kettle
toaster
A
Z
Y
S
2
S
1
20V
X
100 W
150 W
Grade 11 Electric circuits R Toerien Page 14
Power and brightness
The brightness of a light bulb in a circuit is linked to the energy conversion that takes place in the filament
of the light bulb (electrical energy to light (and heat) energy). Power is the energy that is converted per
second and therefore the brightness of a light bulb can be quantified by considering the power dissipated in
the light bulb.
Power is dependent on the potential difference and the current (P = VI). The resistance of a light bulb
usually remains constant since the actual device does not change and it is usually assumed that the
temperature remains constant at the point of measurement. In the following worked examples the changes
in the brightness of light bulbs are discussed when more light bulbs are added in series and parallel.
Light bulbs in series
The circuit on the right has a light bulb L1 connected to a battery.
When a second and third light bulb (L2 and L3) are connected in
series, the brightness of L1 will decrease. When more resistors are
connected in series, the total resistance of the circuit increases.
This causes the current to decrease. Since the potential difference
remains the same, power will decrease (P = VI) and the light bulbs
will be less bright.
Light bulbs in parallel
In the circuit on the right, L2 and L3 are added. When more
identical light bulbs are connected in parallel, the total resistance
of the circuit will decrease. This will cause the total current to
increase. However, the current in each ‘branch’ of the parallel
connection will remain the same since the potential difference
across each light bulb and the resistance of each light bulb did not
change. The brightness of the light bulbs will therefore not be
affected, L1 will remain as bright as it was before L2 and L3 were
added.
In summary, when identical light bulbs are added in series:
- Total current decreases
- Power decreases
- Light bulbs are less bright
When identical light bulbs are added in parallel
- Total current increases
- Current in each branch remains the same
- Power remains the same, therefore the light bulbs will burn with the same brightness as before.
The above is applicable if the resistance of the light bulb in question remains constant.
A
V
L
1
L
2
L
3
A
V
L
2
L
1
L
3
Power and brightness
The brightness of a light bulb in a circuit is linked to the energy conversion that takes place in the filament
of the light bulb (electrical energy to light (and heat) energy). Power is the energy that is converted per
second and therefore the brightness of a light bulb can be quantified by considering the power dissipated in
the light bulb.
Power is dependent on the potential difference and the current (P = VI). The resistance of a light bulb
usually remains constant since the actual device does not change and it is usually assumed that the
temperature remains constant at the point of measurement. In the following worked examples the changes
in the brightness of light bulbs are discussed when more light bulbs are added in series and parallel.
Light bulbs in series
The circuit on the right has a light bulb L1 connected to a battery.
When a second and third light bulb (L2 and L3) are connected in
series, the brightness of L1 will decrease. When more resistors are
connected in series, the total resistance of the circuit increases.
This causes the current to decrease. Since the potential difference
remains the same, power will decrease (P = VI) and the light bulbs
will be less bright.
Light bulbs in parallel
In the circuit on the right, L2 and L3 are added. When more
identical light bulbs are connected in parallel, the total resistance
of the circuit will decrease. This will cause the total current to
increase. However, the current in each ‘branch’ of the parallel
connection will remain the same since the potential difference
across each light bulb and the resistance of each light bulb did not
change. The brightness of the light bulbs will therefore not be
affected, L1 will remain as bright as it was before L2 and L3 were
added.
In summary, when identical light bulbs are added in series:
- Total current decreases
- Power decreases
- Light bulbs are less bright
When identical light bulbs are added in parallel
- Total current increases
- Current in each branch remains the same
- Power remains the same, therefore the light bulbs will burn with the same brightness as before.
The above is applicable if the resistance of the light bulb in question remains constant.
A
V
L
1
L
2
L
3
A
V
L
2
L
1
L
3
Grade 11 Electric circuits R Toerien Page 15
Exercise 5
5.1 What makes a light bulb shine?
5.2 How do you think the current, potential difference and resistance affect the brightness of a light
bulb?
5.3 (For extension) The following five circuits are given. All the cells and light bulbs are identical. The
light bulbs are labelled from A to K. Sort the light bulbs in the circuits from brightest to dimmest.
Some light bulbs may be of the same brightness, use an equal sign in these cases.
A B C
D
E
F
G
H
I
J
K
Exercise 5
5.1 What makes a light bulb shine?
5.2 How do you think the current, potential difference and resistance affect the brightness of a light
bulb?
5.3 (For extension) The following five circuits are given. All the cells and light bulbs are identical. The
light bulbs are labelled from A to K. Sort the light bulbs in the circuits from brightest to dimmest.
Some light bulbs may be of the same brightness, use an equal sign in these cases.
A B C
D
E
F
G
H
I
J
K
Grade 11 Electric circuits R Toerien Page 16
Switches
Switches are placed in circuits to manipulate the flow of current. The following examples discuss the effect
that switches have on circuits.
Worked example 6.1
Consider Circuit 1 on the right. When switch S is open
the current will flow through both resistors. If the resistors
are identical (R1 = R2) then the reading on the voltmeters
will be the same (V1 = V2) and equal to 6V each.
If switch S is closed, no current will flow through R2 as S
provides a short circuit or path of no resistance for the
current. The reading on V2 will be 0 V as no current flows
through R2. The total resistance in the circuit will be less
(RT↓) and the total current (IT↑) will increase. The reading on
the ammeter will therefore increase.
Worked example 6.2
Consider Circuit 2 on the right. When switch S is open, no current
flows in the circuit. The readings on V1 and V2 will both be 0 V since
no current is flowing. The voltmeters will also not measure the emf
of the battery since there is an open switch between them and the
battery. Voltmeter V3 will measure the emf of the battery (12 V).
When the switch is closed, V1 will measure the potential difference
across R1 and V2 will measure the potential difference across R2.
If the resistors are identical, the readings on V1 and V2 will be the same
and equal to 6 V each. V3 will read 0 V since the switch offers no
resistance to the charges flowing through it.
Worked example 6.3
Consider Circuit 3 on the right. When both switches are open, no current
flows in the circuit. All ammeters will read 0 A. The voltmeter V1 will read
the emf of 12 V, and V2 will read 0 V.
If S1 is closed, the current flows through R1, but not R2. Ammeter A1 will
read the total current, the same as ammeter A (A = A1). The reading on A2
will be 0 A since no current is flowing through R2. The reading on V2 will
still be 0 V.
If S1 and S2 are both closed the current will flow through both resistors.
If the resistors are identical, the current will split equally between them,
and be twice as big as before. The reading on A will be double (A = A1 + A2)
as the more resistors there are in parallel, the higher the current will be.
The reading on V1 and V2 will be 12 V.
V
1
R
1
12V
S V
2
R
2
A
V
1
R
1
12V
S
V
2
R
2
V
3
Circuit 1
Circuit 2
A
R
2
R
1
S
2
S
1
V
1
Circuit 3
12V
A
2
A
1
V
2
Switches
Switches are placed in circuits to manipulate the flow of current. The following examples discuss the effect
that switches have on circuits.
Worked example 6.1
Consider Circuit 1 on the right. When switch S is open
the current will flow through both resistors. If the resistors
are identical (R1 = R2) then the reading on the voltmeters
will be the same (V1 = V2) and equal to 6V each.
If switch S is closed, no current will flow through R2 as S
provides a short circuit or path of no resistance for the
current. The reading on V2 will be 0 V as no current flows
through R2. The total resistance in the circuit will be less
(RT↓) and the total current (IT↑) will increase. The reading on
the ammeter will therefore increase.
Worked example 6.2
Consider Circuit 2 on the right. When switch S is open, no current
flows in the circuit. The readings on V1 and V2 will both be 0 V since
no current is flowing. The voltmeters will also not measure the emf
of the battery since there is an open switch between them and the
battery. Voltmeter V3 will measure the emf of the battery (12 V).
When the switch is closed, V1 will measure the potential difference
across R1 and V2 will measure the potential difference across R2.
If the resistors are identical, the readings on V1 and V2 will be the same
and equal to 6 V each. V3 will read 0 V since the switch offers no
resistance to the charges flowing through it.
Worked example 6.3
Consider Circuit 3 on the right. When both switches are open, no current
flows in the circuit. All ammeters will read 0 A. The voltmeter V1 will read
the emf of 12 V, and V2 will read 0 V.
If S1 is closed, the current flows through R1, but not R2. Ammeter A1 will
read the total current, the same as ammeter A (A = A1). The reading on A2
will be 0 A since no current is flowing through R2. The reading on V2 will
still be 0 V.
If S1 and S2 are both closed the current will flow through both resistors.
If the resistors are identical, the current will split equally between them,
and be twice as big as before. The reading on A will be double (A = A1 + A2)
as the more resistors there are in parallel, the higher the current will be.
The reading on V1 and V2 will be 12 V.
V
1
R
1
12V
S V
2
R
2
A
V
1
R
1
12V
S
V
2
R
2
V
3
Circuit 1
Circuit 2
A
R
2
R
1
S
2
S
1
V
1
Circuit 3
12V
A
2
A
1
V
2
Grade 11 Electric circuits R Toerien Page 17
Exercise 6
6.1 In the circuit represented below, the battery has an emf of 12 V and negligible internal resistance.
Voltmeter V1 is connected across the battery and voltmeter V2 is connected across the open
switch S. The resistance of the connecting wires and the ammeter can be ignored.
Switch S is open.
6.1.1 What is the reading on V1?
6.1.2 What is the reading on V2?
Switch S is then closed.
6.1.3 Calculate the effective resistance of the parallel combination of resistors.
6.1.4 Calculate the reading on the ammeter.
6.1.5 Calculate the energy transferred in the 1 Ω resistor in 1,5 minutes.
A
V
2
1 Ω
6 Ω
3 Ω
V
1
emf = 12 V
S
Exercise 6
6.1 In the circuit represented below, the battery has an emf of 12 V and negligible internal resistance.
Voltmeter V1 is connected across the battery and voltmeter V2 is connected across the open
switch S. The resistance of the connecting wires and the ammeter can be ignored.
Switch S is open.
6.1.1 What is the reading on V1?
6.1.2 What is the reading on V2?
Switch S is then closed.
6.1.3 Calculate the effective resistance of the parallel combination of resistors.
6.1.4 Calculate the reading on the ammeter.
6.1.5 Calculate the energy transferred in the 1 Ω resistor in 1,5 minutes.
A
V
2
1 Ω
6 Ω
3 Ω
V
1
emf = 12 V
S
Grade 11 Electric circuits R Toerien Page 18
6.2 In the circuit represented below, the battery has an emf of 24 V. Voltmeter V1 is connected as
indicated and voltmeter V2 is connected across the three parallel resistors. The resistances of the
battery, connecting wires and ammeter can be ignored.
Switch S is open.
6.2.1 What is the reading on V1?
6.2.2 What is the reading on V2?
Switch S is now closed.
6.2.3 Calculate the effective resistance of the entire circuit.
6.2.4 Calculate the reading on the ammeter.
6.2.5 Calculate the amount of charge moving past a cross-section of the 8 Ω resistor in one
minute.
A V
1
8 Ω
6 Ω
3 Ω
V
2
emf = 24 V
S
6 Ω
6.2 In the circuit represented below, the battery has an emf of 24 V. Voltmeter V1 is connected as
indicated and voltmeter V2 is connected across the three parallel resistors. The resistances of the
battery, connecting wires and ammeter can be ignored.
Switch S is open.
6.2.1 What is the reading on V1?
6.2.2 What is the reading on V2?
Switch S is now closed.
6.2.3 Calculate the effective resistance of the entire circuit.
6.2.4 Calculate the reading on the ammeter.
6.2.5 Calculate the amount of charge moving past a cross-section of the 8 Ω resistor in one
minute.
A V
1
8 Ω
6 Ω
3 Ω
V
2
emf = 24 V
S
6 Ω
Grade 11 Electric circuits R Toerien Page 19
Internal resistance
Up to now, we have assumed that batteries have no resistance, but this is not the case in real life. Due to
the material a battery is made of, and the conductivity of the electrolyte, the battery opposes the flow of
charge through it. This is called the internal resistance of the battery and the symbol r is used for it. Internal
resistance can be calculated like any other resistance in a circuit by using the
equation r =
Vint
I
. The internal voltage (Vint) refers to the energy transferred inside
the battery to move charge through the battery and I is the total current in the
circuit. When we indicate internal resistance in a circuit, the symbol on the right is
used. The small resistor shows the resistance inside the battery.
Vint can be observed when a voltmeter is connected across the battery and a
reading is taken without any current flowing in the circuit. This is the emf (E or
Ɛ) of the battery (3V in the diagram). When the switch is closed, the reading on
the voltmeter drops (to 2,8 V in the diagram). It is as if volts have ‘disappeared’,
and this is sometimes referred to as ‘lost volts’. The volts are not really ‘lost’, but
instead, the voltmeter now measures the potential difference across the
external circuit (Vext or Vload) and the difference between the two readings is
equal to Vint. This can be expressed as follows:
emf = Vload + Vinternal resistance
emf = Vext + Vint
emf = IRext + Ir
E = I (R + r)
The emf is the energy provided by a battery per unit charge flowing through the battery.
Worked example 7
A battery with an emf of 30 V, and unknown internal resistance r, is connected to resistors as shown in the
circuit diagram below. Ignore the resistance of the ammeter and the connecting wires. The current passing
through the 10 Ω resistor is 0,6 A.
Calculate the:
7.1 Equivalent resistance of the two resistors in parallel
7.2 Current through the 8 Ω resistor
7.3 Internal resistance of the battery
r
emf = 12 V, r = 1 Ω
R
3V
R
2,8V
A
5 Ω
10 Ω
6 Ω
V
8 Ω
emf = 30 V
r
Internal resistance
Up to now, we have assumed that batteries have no resistance, but this is not the case in real life. Due to
the material a battery is made of, and the conductivity of the electrolyte, the battery opposes the flow of
charge through it. This is called the internal resistance of the battery and the symbol r is used for it. Internal
resistance can be calculated like any other resistance in a circuit by using the
equation r =
Vint
I
. The internal voltage (Vint) refers to the energy transferred inside
the battery to move charge through the battery and I is the total current in the
circuit. When we indicate internal resistance in a circuit, the symbol on the right is
used. The small resistor shows the resistance inside the battery.
Vint can be observed when a voltmeter is connected across the battery and a
reading is taken without any current flowing in the circuit. This is the emf (E or
Ɛ) of the battery (3V in the diagram). When the switch is closed, the reading on
the voltmeter drops (to 2,8 V in the diagram). It is as if volts have ‘disappeared’,
and this is sometimes referred to as ‘lost volts’. The volts are not really ‘lost’, but
instead, the voltmeter now measures the potential difference across the
external circuit (Vext or Vload) and the difference between the two readings is
equal to Vint. This can be expressed as follows:
emf = Vload + Vinternal resistance
emf = Vext + Vint
emf = IRext + Ir
E = I (R + r)
The emf is the energy provided by a battery per unit charge flowing through the battery.
Worked example 7
A battery with an emf of 30 V, and unknown internal resistance r, is connected to resistors as shown in the
circuit diagram below. Ignore the resistance of the ammeter and the connecting wires. The current passing
through the 10 Ω resistor is 0,6 A.
Calculate the:
7.1 Equivalent resistance of the two resistors in parallel
7.2 Current through the 8 Ω resistor
7.3 Internal resistance of the battery
r
emf = 12 V, r = 1 Ω
R
3V
R
2,8V
A
5 Ω
10 Ω
6 Ω
V
8 Ω
emf = 30 V
r
Grade 11 Electric circuits R Toerien Page 20
Exercise 7
7.1 In the circuit shown, two 60 Ω resistors connected in parallel are connected in series with a 25 Ω
resistor. The battery has an emf of 12 V and an internal resistance of 1,5 Ω.
Calculate the:
7.1.1 Equivalent resistance of the parallel combination
7.1.2 Total current in the circuit
7.1.3 Potential difference across the parallel resistors
7.2 In the circuit shown, the reading on ammeter A is 0,2 A. The battery has an emf of 9 V and internal
resistance r.
7.2.1 Calculate the current through the 5,5 Ω resistor.
7.2.2 Calculate the internal resistance of the battery.
emf = 12 V
1,5Ω
60 Ω
60 Ω
25 Ω
5,5 Ω
11 Ω
emf = 9 V
r
11 Ω
S
A
Exercise 7
7.1 In the circuit shown, two 60 Ω resistors connected in parallel are connected in series with a 25 Ω
resistor. The battery has an emf of 12 V and an internal resistance of 1,5 Ω.
Calculate the:
7.1.1 Equivalent resistance of the parallel combination
7.1.2 Total current in the circuit
7.1.3 Potential difference across the parallel resistors
7.2 In the circuit shown, the reading on ammeter A is 0,2 A. The battery has an emf of 9 V and internal
resistance r.
7.2.1 Calculate the current through the 5,5 Ω resistor.
7.2.2 Calculate the internal resistance of the battery.
emf = 12 V
1,5Ω
60 Ω
60 Ω
25 Ω
5,5 Ω
11 Ω
emf = 9 V
r
11 Ω
S
A
Grade 11 Electric circuits R Toerien Page 21
7.3 The headlights of a car are connected in parallel to a 12 V battery, as
shown in the simplified circuit diagram on the right. The internal
resistance of the battery is 0,1 Ω and each headlight has a resistance
of 1,4 Ω. The starter motor is connected in parallel with the headlights
and controlled by the ignition switch, S2. The resistance of the
connecting wires may be ignored.
With only switch S1 closed, calculate the following:
7.3.1 The effective resistance of the two headlights
7.3.2 The potential difference across the two headlights
7.3.3 The power dissipated by one of the headlights
7.3.4 Ignition switch S2 is now closed (whilst S1 is also closed) for a
short time and the starter motor, with VERY LOW
RESISTANCE, rotates. How will the brightness of the headlights
be affected while switch S2 is closed? Write down only
INCREASES, DECREASES , or REMAINS THE SAME. Fully
explain how you arrived at the answer.
S
1
S
2
0,1 Ω
emf = 12 V
Starter motor
1,4 Ω
1,4 Ω
7.3 The headlights of a car are connected in parallel to a 12 V battery, as
shown in the simplified circuit diagram on the right. The internal
resistance of the battery is 0,1 Ω and each headlight has a resistance
of 1,4 Ω. The starter motor is connected in parallel with the headlights
and controlled by the ignition switch, S2. The resistance of the
connecting wires may be ignored.
With only switch S1 closed, calculate the following:
7.3.1 The effective resistance of the two headlights
7.3.2 The potential difference across the two headlights
7.3.3 The power dissipated by one of the headlights
7.3.4 Ignition switch S2 is now closed (whilst S1 is also closed) for a
short time and the starter motor, with VERY LOW
RESISTANCE, rotates. How will the brightness of the headlights
be affected while switch S2 is closed? Write down only
INCREASES, DECREASES , or REMAINS THE SAME. Fully
explain how you arrived at the answer.
S
1
S
2
0,1 Ω
emf = 12 V
Starter motor
1,4 Ω
1,4 Ω
Grade 11 Electric circuits R Toerien Page 22
Explaining ‘lost volts’
When resistors are added to a circuit (in parallel or in series), the current is affected. This will, in turn, affect
the ‘lost volts’ as Vint = Ir. The emf of a battery is a physical property of the battery and will remain constant.
Since emf = Vext + Vint, the potential difference of the external circuit will be affected. This is discussed in
the worked example below.
Worked example 8
A cell of unknown internal resistance, r, has an emf (Ɛ) of 1,5 V. It is connected in a circuit to three
resistors, a high-resistance voltmeter, low-resistance ammeters, and a switch S as shown below.
When switch S is closed, the voltmeter reads 1,36 V.
8.1 Calculate the reading on A2.
8.2 Determine the internal resistance of the cell.
8.3 An additional resistor X is connected in parallel to the
3 Ω resistor in the circuit. Explain in detail how this
will affect the ammeter and voltmeter readings?
8.4 If resistor X is connected in series to the parallel
combination, how will the ammeter and voltmeter
reading be affected?
8.3 Adding an additional resistor in parallel will decrease the total resistance of the external circuit. This
will result in an increase in the total current in the circuit (RT↓ therefore IT↑). The ammeter reading
will increase.
Since the internal resistance r remains constant and the total current increased, the ‘lost volts’ in the
battery will increase (Vint = Ir and I↑ therefore Vint↑). Since the emf remains constant, the potential
difference in the external circuit (Vext) will decrease. (emf = Vext + Vint; Vint↑ therefore Vext↓).
The voltmeter reading will decrease.
8.4 Adding an additional resistor in series will increase the total resistance of the external circuit. This
will result in a decrease in the total current in the circuit (RT↑ therefore IT↓). The reading on the
ammeter will therefore decrease.
The ‘lost volts’ will decrease (Vint = Ir and if IT↓ then Vint↓) and since the emf remains constant, the
potential difference in the external circuit (Vext) will increase (emf = Vext + Vint ; Vint↓ therefore Vext↑).
The reading on the voltmeter will increase.
S
V
emf = 1,5 V
3 Ω
2 Ω 4 Ω
A1
r
A2
Explaining ‘lost volts’
When resistors are added to a circuit (in parallel or in series), the current is affected. This will, in turn, affect
the ‘lost volts’ as Vint = Ir. The emf of a battery is a physical property of the battery and will remain constant.
Since emf = Vext + Vint, the potential difference of the external circuit will be affected. This is discussed in
the worked example below.
Worked example 8
A cell of unknown internal resistance, r, has an emf (Ɛ) of 1,5 V. It is connected in a circuit to three
resistors, a high-resistance voltmeter, low-resistance ammeters, and a switch S as shown below.
When switch S is closed, the voltmeter reads 1,36 V.
8.1 Calculate the reading on A2.
8.2 Determine the internal resistance of the cell.
8.3 An additional resistor X is connected in parallel to the
3 Ω resistor in the circuit. Explain in detail how this
will affect the ammeter and voltmeter readings?
8.4 If resistor X is connected in series to the parallel
combination, how will the ammeter and voltmeter
reading be affected?
8.3 Adding an additional resistor in parallel will decrease the total resistance of the external circuit. This
will result in an increase in the total current in the circuit (RT↓ therefore IT↑). The ammeter reading
will increase.
Since the internal resistance r remains constant and the total current increased, the ‘lost volts’ in the
battery will increase (Vint = Ir and I↑ therefore Vint↑). Since the emf remains constant, the potential
difference in the external circuit (Vext) will decrease. (emf = Vext + Vint; Vint↑ therefore Vext↓).
The voltmeter reading will decrease.
8.4 Adding an additional resistor in series will increase the total resistance of the external circuit. This
will result in a decrease in the total current in the circuit (RT↑ therefore IT↓). The reading on the
ammeter will therefore decrease.
The ‘lost volts’ will decrease (Vint = Ir and if IT↓ then Vint↓) and since the emf remains constant, the
potential difference in the external circuit (Vext) will increase (emf = Vext + Vint ; Vint↓ therefore Vext↑).
The reading on the voltmeter will increase.
S
V
emf = 1,5 V
3 Ω
2 Ω 4 Ω
A1
r
A2
Grade 11 Electric circuits R Toerien Page 23
Exercise 8
8.1 Two identical cells, each with an emf of 1,5 V and an internal resistance r, are connected in series
with each other and to the resistors as shown in the diagram.
8.1.1 Write down the total emf of the circuit.
When switch S is closed, the potential difference across the 4 Ω resistor is 2,8 V.
8.1.2 Calculate the total current in the circuit.
8.1.3 Calculate the internal resistance r of EACH cell.
8.1.4 An unknown resistor is now connected in parallel with the 4 Ω and 1 Ω resistors.
How will this change affect the magnitude of:
8.1.4a The internal resistance of the battery. Write down only INCREASES, DECREASES,
or REMAINS THE SAME.
8.1.4b The reading on the voltmeter. Write down only INCREASES, DECREASES , or
REMAINS THE SAME. Explain the answer by referring to resistance, current, and
'lost volts'.
S
V
4 Ω
1 Ω
r r
Exercise 8
8.1 Two identical cells, each with an emf of 1,5 V and an internal resistance r, are connected in series
with each other and to the resistors as shown in the diagram.
8.1.1 Write down the total emf of the circuit.
When switch S is closed, the potential difference across the 4 Ω resistor is 2,8 V.
8.1.2 Calculate the total current in the circuit.
8.1.3 Calculate the internal resistance r of EACH cell.
8.1.4 An unknown resistor is now connected in parallel with the 4 Ω and 1 Ω resistors.
How will this change affect the magnitude of:
8.1.4a The internal resistance of the battery. Write down only INCREASES, DECREASES,
or REMAINS THE SAME.
8.1.4b The reading on the voltmeter. Write down only INCREASES, DECREASES , or
REMAINS THE SAME. Explain the answer by referring to resistance, current, and
'lost volts'.
S
V
4 Ω
1 Ω
r r
Grade 11 Electric circuits R Toerien Page 24
8.2 Three resistors, R1, R2 and R3, are connected to a battery, as shown in the circuit diagram below.
The internal resistance of the battery is 0,3 Ω. The resistance of R2 and R3 is equal. The resistance
of R1 is half that of R2. When both switches are open, the voltmeter across the battery reads 9 V.
8.2.1 What is the value of the emf of the battery? Give a reason for the answer.
8.2.2 When only switch S1 is closed, the reading on the ammeter is 3 A. Calculate the resistance
of R1.
Both switches S1 and S2 are now closed.
8.2.3 How will the resistance of the circuit change? Write down only INCREASES, DECREASES,
or REMAINS THE SAME.
8.2.4 A conducting wire of negligible resistance is connected between points Q and N. What effect
will this have on the 'lost volts'? Explain the answer.
N
0,3Ω
A
V
1
R
1
= R
R
2
= 2R
9V
S
2
R
3
= 2R
Q
S
1
r
8.2 Three resistors, R1, R2 and R3, are connected to a battery, as shown in the circuit diagram below.
The internal resistance of the battery is 0,3 Ω. The resistance of R2 and R3 is equal. The resistance
of R1 is half that of R2. When both switches are open, the voltmeter across the battery reads 9 V.
8.2.1 What is the value of the emf of the battery? Give a reason for the answer.
8.2.2 When only switch S1 is closed, the reading on the ammeter is 3 A. Calculate the resistance
of R1.
Both switches S1 and S2 are now closed.
8.2.3 How will the resistance of the circuit change? Write down only INCREASES, DECREASES,
or REMAINS THE SAME.
8.2.4 A conducting wire of negligible resistance is connected between points Q and N. What effect
will this have on the 'lost volts'? Explain the answer.
N
0,3Ω
A
V
1
R
1
= R
R
2
= 2R
9V
S
2
R
3
= 2R
Q
S
1
r
Grade 11 Electric circuits R Toerien Page 25
8.3 In the two circuits represented below, the
same battery is used. The emf of the battery
and its internal resistance, r, are unknown.
When three resistors, 4 Ω, 10 Ω and 20 Ω
are connected in parallel, as shown in
Figure 1, the voltmeter reads 10 V. The
parallel combination is disconnected and
instead a resistor of 5,5 Ω is connected in
series with the same battery as shown in
Figure 2. The voltmeter then reads 11 V.
8.3.1 Calculate the effective resistance of the three external resistors in Figure 1.
8.3.2 Explain why the voltmeter reading in Figure 2 is greater than that in Figure 1.
8.3.3 Calculate the reading on ammeter A1 in Figure 1.
8.3.4 Calculate the reading on ammeter A2 in Figure 2.
8.3.5 Calculate the internal resistance, r, of the battery.
8.3.6 Calculate the emf of the battery.
Figure 1
A
1
10V
20 Ω
10 Ω
r
4 Ω
Figure 2
A
2
11V
5,5 Ω
r
8.3 In the two circuits represented below, the
same battery is used. The emf of the battery
and its internal resistance, r, are unknown.
When three resistors, 4 Ω, 10 Ω and 20 Ω
are connected in parallel, as shown in
Figure 1, the voltmeter reads 10 V. The
parallel combination is disconnected and
instead a resistor of 5,5 Ω is connected in
series with the same battery as shown in
Figure 2. The voltmeter then reads 11 V.
8.3.1 Calculate the effective resistance of the three external resistors in Figure 1.
8.3.2 Explain why the voltmeter reading in Figure 2 is greater than that in Figure 1.
8.3.3 Calculate the reading on ammeter A1 in Figure 1.
8.3.4 Calculate the reading on ammeter A2 in Figure 2.
8.3.5 Calculate the internal resistance, r, of the battery.
8.3.6 Calculate the emf of the battery.
Figure 1
A
1
10V
20 Ω
10 Ω
r
4 Ω
Figure 2
A
2
11V
5,5 Ω
r
Grade 11 Electric circuits R Toerien Page 26
Experimental determination of the internal resistance and emf of a battery
Different experiments can be done to determine the emf and internal resistance of a battery. Below are
three different experimental setups, using similar equipment, but drawing different graphs to determine the
internal resistance and the emf of a battery.
Experiment 1: Finding the internal resistance and emf of a battery
Learners set up a circuit according to the circuit diagram below to determine the internal resistance and emf
of a battery. The rheostat (variable resistor) was adjusted to ensure different readings for the total current in
the circuit. The potential difference was recorded for each current reading as shown in the table below.
Rheostat
adjustment
Current (A) Potential
difference (V)
1 0,3 2,80
2 0,4 2,68
3 0,5 2,60
4 0,6 2,50
5 0,7 2,42
6 0,8 2,30
In this experiment, the current was controlled (independent variable) by manipulating the rheostat. The
resistance of the rheostat started high so that the current is low. The resistance of the rheostat is then
decreased so that the current will increase by 0,1 A each time. In between each reading, the switch is
opened to prevent the circuit from heating up. This is done because temperature affects resistance. As the
total resistance decreased (RT↓) the total current increased (IT↑). The voltmeter reading decreased each
time. The voltmeter measured the potential difference of the external circuit (Vext) which is the dependent
variable. A graph of V versus I is drawn as shown below.
The y-intercept and gradient are used to determine the emf and internal resistance.
3,1
3,0
2,9
2,8
2,7
2,6
2,5
2,4
2,3
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
Current strength I (A)
Potential difference V (V)
Graph of potential difference (V) versus current strength (I)
Gradient =
∆V
∆I
=
2,3 – 2,8
0,8 – 0,3
=
– 0,5 V
0,5 A
= –1 Ω
gradient (m) = –r
∴ internal resistance (r) = 1 Ω
y-intercept = 3,1 V
y-intercept (c) = emf
∴ emf = 3,1 V
A
V
r
Experimental determination of the internal resistance and emf of a battery
Different experiments can be done to determine the emf and internal resistance of a battery. Below are
three different experimental setups, using similar equipment, but drawing different graphs to determine the
internal resistance and the emf of a battery.
Experiment 1: Finding the internal resistance and emf of a battery
Learners set up a circuit according to the circuit diagram below to determine the internal resistance and emf
of a battery. The rheostat (variable resistor) was adjusted to ensure different readings for the total current in
the circuit. The potential difference was recorded for each current reading as shown in the table below.
Rheostat
adjustment
Current (A) Potential
difference (V)
1 0,3 2,80
2 0,4 2,68
3 0,5 2,60
4 0,6 2,50
5 0,7 2,42
6 0,8 2,30
In this experiment, the current was controlled (independent variable) by manipulating the rheostat. The
resistance of the rheostat started high so that the current is low. The resistance of the rheostat is then
decreased so that the current will increase by 0,1 A each time. In between each reading, the switch is
opened to prevent the circuit from heating up. This is done because temperature affects resistance. As the
total resistance decreased (RT↓) the total current increased (IT↑). The voltmeter reading decreased each
time. The voltmeter measured the potential difference of the external circuit (Vext) which is the dependent
variable. A graph of V versus I is drawn as shown below.
The y-intercept and gradient are used to determine the emf and internal resistance.
3,1
3,0
2,9
2,8
2,7
2,6
2,5
2,4
2,3
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
Current strength I (A)
Potential difference V (V)
Graph of potential difference (V) versus current strength (I)
Gradient =
∆V
∆I
=
2,3 – 2,8
0,8 – 0,3
=
– 0,5 V
0,5 A
= –1 Ω
gradient (m) = –r
∴ internal resistance (r) = 1 Ω
y-intercept = 3,1 V
y-intercept (c) = emf
∴ emf = 3,1 V
A
V
r
Grade 11 Electric circuits R Toerien Page 27
Experiment 1 (cont.)
In this experiment,
• the current is varied (by adjusting the rheostat) (independent variable) and
• the potential difference of the external circuit is measured (dependent variable).
A graph of V versus I is drawn
Experiment 2
In this experiment,
• the external resistance of the circuit is varied (independent variable) and
• the potential difference of the external circuit is measured (dependent
variable).
A graph of
1
V
versus
1
R
is drawn.
Experiment 3
In this experiment,
• the current is varied by adjusting the rheostat and the total current I is measured (independent variable)
• the resistance of the external circuit is measured (dependent variable) for each current value.
1
V
1
E
1
R
gradient =
r
E
1
V
=
r
ER
+
1
E
1
V
=
r
E
×
1
R
+
1
E
y = m?????? + c
gradient =
r
E
y-intercept =
1
E
Deriving
1
V
=
r
ER
+
1
E
:
E = I (R + r)
× R : ER = IR (R +r)
ER = V (R + r)
1
V
=
R + r
ER
1
V
=
R
ER
+
r
ER
1
V
=
R
ER
+
r
ER
1
V
=
r
ER
+
1
E
Not for exam purposes!
R
–r
1
I
gradient = E
R =
E
I
– r
R =
E
1
×
1
I
– r
y = m?????? + c
gradient = E
y-intercept = – r
Deriving R =
E
I
– r :
E = I (R + r)
E
I
= R + r
R =
E
I
– r
Not for exam purposes!
A
V
r
V
E
I
gradient = –r
E = I (R + r)
E = IR + Ir
IR = – Ir + E
V = –rI + E
y = m?????? + c
Experiment 1 (cont.)
In this experiment,
• the current is varied (by adjusting the rheostat) (independent variable) and
• the potential difference of the external circuit is measured (dependent variable).
A graph of V versus I is drawn
Experiment 2
In this experiment,
• the external resistance of the circuit is varied (independent variable) and
• the potential difference of the external circuit is measured (dependent
variable).
A graph of
1
V
versus
1
R
is drawn.
Experiment 3
In this experiment,
• the current is varied by adjusting the rheostat and the total current I is measured (independent variable)
• the resistance of the external circuit is measured (dependent variable) for each current value.
1
V
1
E
1
R
gradient =
r
E
1
V
=
r
ER
+
1
E
1
V
=
r
E
×
1
R
+
1
E
y = m?????? + c
gradient =
r
E
y-intercept =
1
E
Deriving
1
V
=
r
ER
+
1
E
:
E = I (R + r)
× R : ER = IR (R +r)
ER = V (R + r)
1
V
=
R + r
ER
1
V
=
R
ER
+
r
ER
1
V
=
R
ER
+
r
ER
1
V
=
r
ER
+
1
E
Not for exam purposes!
R
–r
1
I
gradient = E
R =
E
I
– r
R =
E
1
×
1
I
– r
y = m?????? + c
gradient = E
y-intercept = – r
Deriving R =
E
I
– r :
E = I (R + r)
E
I
= R + r
R =
E
I
– r
Not for exam purposes!
A
V
r
V
E
I
gradient = –r
E = I (R + r)
E = IR + Ir
IR = – Ir + E
V = –rI + E
y = m?????? + c
Grade 11 Electric circuits R Toerien Page 28
Exercise 9
9.1 Learners in a school are required to perform an experiment to
obtain the relationship between the potential difference across
an unknown resistor and the current in the resistor. Their results
are shown in the table.
9.1.1 What is the investigative question for this investigation?
9.1.2 Write down a possible hypothesis for this investigation.
9.1.3 Write down ONE variable that the learners must control during this investigation.
9.1.4 List ALL the apparatus that the learners will need for this experiment.
9.1.5 Draw a circuit diagram that they can use to assemble the apparatus.
9.1.6 Plot the graph of potential difference (y-axis) versus current (x-axis).
9.1.7 State the mathematical relationship, in words, between the potential difference across the
unknown resistor and the current in the resistor.
9.1.8 Calculate the gradient of the line.
9.1.9 What physical quantity does the gradient of this line represent?
9.1.10 This experiment forms the basis of an important law. Name and state this law.
Potential
difference in
volt (V)
Current (I) in
ampere (A)
0 0
1,5 0,26
3,2 0,60
4,2 0,72
5,4 0,96
Exercise 9
9.1 Learners in a school are required to perform an experiment to
obtain the relationship between the potential difference across
an unknown resistor and the current in the resistor. Their results
are shown in the table.
9.1.1 What is the investigative question for this investigation?
9.1.2 Write down a possible hypothesis for this investigation.
9.1.3 Write down ONE variable that the learners must control during this investigation.
9.1.4 List ALL the apparatus that the learners will need for this experiment.
9.1.5 Draw a circuit diagram that they can use to assemble the apparatus.
9.1.6 Plot the graph of potential difference (y-axis) versus current (x-axis).
9.1.7 State the mathematical relationship, in words, between the potential difference across the
unknown resistor and the current in the resistor.
9.1.8 Calculate the gradient of the line.
9.1.9 What physical quantity does the gradient of this line represent?
9.1.10 This experiment forms the basis of an important law. Name and state this law.
Potential
difference in
volt (V)
Current (I) in
ampere (A)
0 0
1,5 0,26
3,2 0,60
4,2 0,72
5,4 0,96
Grade 11 Electric circuits R Toerien Page 29
9.2 A group of learners conducts an experiment to determine the emf (Ɛ) and internal resistance (r) of a
battery. They connect a battery to a rheostat (variable resistor), a low-resistance ammeter, and a
high-resistance voltmeter as shown in the diagram below. The data they obtained is shown in the
table below.
9.2.1 State ONE factor which must be kept constant during the experiment.
9.2.2 Use the information in the table to plot a graph of potential difference versus current and draw
the line of best fit.
Use the graph drawn to determine the following:
9.2.3 Emf (Ɛ) of the battery
9.2.4 Internal resistance of the battery, without using any form of the equation Ɛ = I(R + r)
Ammeter reading (A) Voltmeter reading (V)
0,58 2,0
0,46 3,0
0,36 4,0
0,24 5,0
0,14 6,0
A
V
r
9.2 A group of learners conducts an experiment to determine the emf (Ɛ) and internal resistance (r) of a
battery. They connect a battery to a rheostat (variable resistor), a low-resistance ammeter, and a
high-resistance voltmeter as shown in the diagram below. The data they obtained is shown in the
table below.
9.2.1 State ONE factor which must be kept constant during the experiment.
9.2.2 Use the information in the table to plot a graph of potential difference versus current and draw
the line of best fit.
Use the graph drawn to determine the following:
9.2.3 Emf (Ɛ) of the battery
9.2.4 Internal resistance of the battery, without using any form of the equation Ɛ = I(R + r)
Ammeter reading (A) Voltmeter reading (V)
0,58 2,0
0,46 3,0
0,36 4,0
0,24 5,0
0,14 6,0
A
V
r
Grade 11 Electric circuits R Toerien Page 30
9.3 In an experiment, learners use a circuit to
determine the internal resistance of a cell. The
circuit consists of a cell of emf E and internal
resistance r. A voltmeter is placed across a
variable resistor which can be set to known
values of R. The equation used by the learners
is:
1
V
=
r
ER
+
1
E
.
They obtain the graph shown on the right.
9.3.1 Write down a mathematical relationship
for the slope of the graph.
Use the information in the graph to determine:
9.3.2 The emf of the cell
9.3.3 The internal resistance of the cell
9.4 Learners perform an experiment to determine
the emf (Ɛ) and the internal resistance (r) of a
battery. They use their recorded readings of
current and resistance, together with the
equation R =
Ɛ
I
– r, to obtain the graph
shown on the right.
9.4.1 Which variable has to be kept constant
in the experiment? (1)
Refer to the graph for the following questions.
9.4.2 Write down the value of the internal
resistance of the cell. (2)
9.4.3 Calculate the emf of the battery. (3)
[6]
15
10
5
0
0,5 1,0 1,5
Resistance (
Ω
)
Graph of resistance versus
1
current
–5
1
current
(A
–1
)
2,0
2,0
1,5
1,0
0,5
1,0 2,0 3,0
1
R
(Ω
–1
)
1
V
(V
–1
)
0,0
Graph of
1
V
versus
1
R
4,0
9.3 In an experiment, learners use a circuit to
determine the internal resistance of a cell. The
circuit consists of a cell of emf E and internal
resistance r. A voltmeter is placed across a
variable resistor which can be set to known
values of R. The equation used by the learners
is:
1
V
=
r
ER
+
1
E
.
They obtain the graph shown on the right.
9.3.1 Write down a mathematical relationship
for the slope of the graph.
Use the information in the graph to determine:
9.3.2 The emf of the cell
9.3.3 The internal resistance of the cell
9.4 Learners perform an experiment to determine
the emf (Ɛ) and the internal resistance (r) of a
battery. They use their recorded readings of
current and resistance, together with the
equation R =
Ɛ
I
– r, to obtain the graph
shown on the right.
9.4.1 Which variable has to be kept constant
in the experiment? (1)
Refer to the graph for the following questions.
9.4.2 Write down the value of the internal
resistance of the cell. (2)
9.4.3 Calculate the emf of the battery. (3)
[6]
15
10
5
0
0,5 1,0 1,5
Resistance (
Ω
)
Graph of resistance versus
1
current
–5
1
current
(A
–1
)
2,0
2,0
1,5
1,0
0,5
1,0 2,0 3,0
1
R
(Ω
–1
)
1
V
(V
–1
)
0,0
Graph of
1
V
versus
1
R
4,0
Grade 11 Electric circuits R Toerien Page 31
9.5 Learners conduct an investigation to verify Ohm’s law. They measure the current through a
conducting wire for different potential differences across its ends. The results obtained are shown in
the graph below.
9.5.1 Which ONE of the measured quantities is the dependent variable?
9.5.2 The graph deviates from Ohm’s law at some point.
9.5.2a Write down the coordinates of the plotted point on the graph beyond which Ohm’s law
is not obeyed.
9.5.2b Give a possible reason for the deviation from Ohm’s law as shown in the graph.
Assume that all measurements are correct.
9.5.3 Calculate the gradient of the line for the section where Ohm’s law is obeyed. Use this to
calculate the resistance of the conducting wire.
0,8
0,6
0,4
0,2
2,0 4,0 6,0
Potential difference V (V)
Current
I
(A)
0,0
Graph of I versus V
0,7
0,5
0,3
0,1
0,9
1,0 3,0 5,0 7,0
9.5 Learners conduct an investigation to verify Ohm’s law. They measure the current through a
conducting wire for different potential differences across its ends. The results obtained are shown in
the graph below.
9.5.1 Which ONE of the measured quantities is the dependent variable?
9.5.2 The graph deviates from Ohm’s law at some point.
9.5.2a Write down the coordinates of the plotted point on the graph beyond which Ohm’s law
is not obeyed.
9.5.2b Give a possible reason for the deviation from Ohm’s law as shown in the graph.
Assume that all measurements are correct.
9.5.3 Calculate the gradient of the line for the section where Ohm’s law is obeyed. Use this to
calculate the resistance of the conducting wire.
0,8
0,6
0,4
0,2
2,0 4,0 6,0
Potential difference V (V)
Current
I
(A)
0,0
Graph of I versus V
0,7
0,5
0,3
0,1
0,9
1,0 3,0 5,0 7,0
Grade 11 Electric circuits R Toerien Page 32
Grade 11 Physics Module P5 Worksheet Electric circuits
QUESTION 1
1.1 Which ONE of the graphs below best represents the relationship between the potential difference (V)
and current (I) for an ohmic conductor? (DBE, Nov 2019)
(2)
1.2 In the circuit diagrams below, the cells and resistors are identical. The cells have negligible internal
resistances.
The power dissipated in resistor X is P. The power dissipated in resistor Y is …
A ¼ P
B ½ P
C 2 P
D 4 P (2)
(DBE, Nov 2018)
1.3 Learners investigate the relationship between current (I) and potential difference (V) at a constant
temperature for three different resistors, X, Y, and Z. They obtain the graphs shown in the right.
The resistances of X, Y, and Z are RX, RY, and RZ respectively.
Which ONE of the following conclusions regarding the resistances of the resistors is CORRECT?
A RX < RY < RZ
B RX = RY = RZ
C RX > RY > RZ
D RX > RY and RY < RZ (2)
(DBE, Nov 2016)
V
I
A
V
I
B
V
I
C
X Y
I (A)
V (V)
X
Y
Z
V
I
D
Grade 11 Physics Module P5 Worksheet Electric circuits
QUESTION 1
1.1 Which ONE of the graphs below best represents the relationship between the potential difference (V)
and current (I) for an ohmic conductor? (DBE, Nov 2019)
(2)
1.2 In the circuit diagrams below, the cells and resistors are identical. The cells have negligible internal
resistances.
The power dissipated in resistor X is P. The power dissipated in resistor Y is …
A ¼ P
B ½ P
C 2 P
D 4 P (2)
(DBE, Nov 2018)
1.3 Learners investigate the relationship between current (I) and potential difference (V) at a constant
temperature for three different resistors, X, Y, and Z. They obtain the graphs shown in the right.
The resistances of X, Y, and Z are RX, RY, and RZ respectively.
Which ONE of the following conclusions regarding the resistances of the resistors is CORRECT?
A RX < RY < RZ
B RX = RY = RZ
C RX > RY > RZ
D RX > RY and RY < RZ (2)
(DBE, Nov 2016)
V
I
A
V
I
B
V
I
C
X Y
I (A)
V (V)
X
Y
Z
V
I
D
Grade 11 Electric circuits R Toerien Page 33
1.4 Four identical bulbs, P, Q, R, and S, are connected to a cell in a circuit, as shown below. The cell has
negligible resistance.
Which ONE of the following statements about the brightness of bulbs, P, Q, R, and S is CORRECT?
A P burns brighter than R.
B S and Q burn brighter than P and R.
C P and R burn brighter than S and Q.
D ALL the bulbs burn with the same brightness. (2)
(DBE, March 2018)
1.5 In the circuit diagram below, the resistance of resistor R1 is TWICE the resistance of resistor R2. The
two resistors are connected in series and identical high-resistance voltmeters are connected across
each resistor.
Which ONE of the following statements concerning the voltmeter readings is CORRECT?
A V1 = 2 V2
B V1 = ½ V2
C V1 = ¼ V2
D 2 V1 = V2 (2)
(DBE, March 2017) [10]
P
S
R
Q
V
1
R
1
V
2
R
2
1.4 Four identical bulbs, P, Q, R, and S, are connected to a cell in a circuit, as shown below. The cell has
negligible resistance.
Which ONE of the following statements about the brightness of bulbs, P, Q, R, and S is CORRECT?
A P burns brighter than R.
B S and Q burn brighter than P and R.
C P and R burn brighter than S and Q.
D ALL the bulbs burn with the same brightness. (2)
(DBE, March 2018)
1.5 In the circuit diagram below, the resistance of resistor R1 is TWICE the resistance of resistor R2. The
two resistors are connected in series and identical high-resistance voltmeters are connected across
each resistor.
Which ONE of the following statements concerning the voltmeter readings is CORRECT?
A V1 = 2 V2
B V1 = ½ V2
C V1 = ¼ V2
D 2 V1 = V2 (2)
(DBE, March 2017) [10]
P
S
R
Q
V
1
R
1
V
2
R
2
Grade 11 Electric circuits R Toerien Page 34
QUESTION 2
In the circuit diagram shown, the battery has an emf of 12 V and an
internal resistance of 0,8 Ω. The resistance of the ammeter and
connecting wires may be ignored.
2.1 Define, in words, the term electromotive force (emf). (2)
2.2 Calculate the effective resistance of the external circuit. (4)
2.3 Calculate the reading on the ammeter. (3)
2.4 Calculate the reading on the voltmeter. (4)
[13]
emf = 12 V
0,8Ω
4 Ω
8 Ω
8 Ω
2 Ω
A
V
QUESTION 2
In the circuit diagram shown, the battery has an emf of 12 V and an
internal resistance of 0,8 Ω. The resistance of the ammeter and
connecting wires may be ignored.
2.1 Define, in words, the term electromotive force (emf). (2)
2.2 Calculate the effective resistance of the external circuit. (4)
2.3 Calculate the reading on the ammeter. (3)
2.4 Calculate the reading on the voltmeter. (4)
[13]
emf = 12 V
0,8Ω
4 Ω
8 Ω
8 Ω
2 Ω
A
V
Grade 11 Electric circuits R Toerien Page 35
QUESTION 3
The battery in the circuit diagram has an emf of 12 V and an unknown internal resistance r. Voltmeter V1 is
connected across the battery and voltmeter V2 is connected across the switch S. The resistance of the
connecting wires and the ammeter is negligible.
3.1 Define the term internal resistance. (2)
3.2 Write down the respective readings on
voltmeters V1 and V2 when switch S is open. (2)
Switch S is now closed. The reading on
voltmeter V1 changes to 9 V.
3.3 What will the new reading on V2 be? (1)
3.4 Calculate the total external resistance of the
circuit. (4)
3.5 Calculate the internal resistance, r, of the
battery. (5)
[14]
A
2 Ω
12 Ω
6 Ω
S
V
2
Ɛ = 12 V
r
V
1
QUESTION 3
The battery in the circuit diagram has an emf of 12 V and an unknown internal resistance r. Voltmeter V1 is
connected across the battery and voltmeter V2 is connected across the switch S. The resistance of the
connecting wires and the ammeter is negligible.
3.1 Define the term internal resistance. (2)
3.2 Write down the respective readings on
voltmeters V1 and V2 when switch S is open. (2)
Switch S is now closed. The reading on
voltmeter V1 changes to 9 V.
3.3 What will the new reading on V2 be? (1)
3.4 Calculate the total external resistance of the
circuit. (4)
3.5 Calculate the internal resistance, r, of the
battery. (5)
[14]
A
2 Ω
12 Ω
6 Ω
S
V
2
Ɛ = 12 V
r
V
1
Grade 11 Electric circuits R Toerien Page 36
QUESTION 4
The circuit diagram shows a battery, with an internal resistance r, connected to three resistors, M, N, and
Y. The resistance of N is 2 Ω and the reading on voltmeter V is 14 V. The reading on ammeter A1 is 2 A
and the reading on ammeter A2 is 1 A. The resistance of the ammeters and the connecting wires may be
ignored.
4.1 State Ohm's law in words. (2)
4.2 How does the resistance of M compare with that
of N? Explain how you arrived at the answer. (2)
4.3 If the emf of the battery is 17 V, calculate the
internal resistance of the battery. (4)
4.4 Calculate the potential difference across
resistor N. (3)
4.5 Calculate the resistance of Y. (4)
[15]
1 A
Y
M N
r
V
A
2
A
1
2 A
14 V
QUESTION 4
The circuit diagram shows a battery, with an internal resistance r, connected to three resistors, M, N, and
Y. The resistance of N is 2 Ω and the reading on voltmeter V is 14 V. The reading on ammeter A1 is 2 A
and the reading on ammeter A2 is 1 A. The resistance of the ammeters and the connecting wires may be
ignored.
4.1 State Ohm's law in words. (2)
4.2 How does the resistance of M compare with that
of N? Explain how you arrived at the answer. (2)
4.3 If the emf of the battery is 17 V, calculate the
internal resistance of the battery. (4)
4.4 Calculate the potential difference across
resistor N. (3)
4.5 Calculate the resistance of Y. (4)
[15]
1 A
Y
M N
r
V
A
2
A
1
2 A
14 V
Grade 11 Electric circuits R Toerien Page 37
QUESTION 5
A learner wants to use a 12 V battery with an internal resistance of 1 Ω to operate an electrical device. He
uses the circuit below to obtain the desired potential difference for the device to function. The resistance of
the device is 5 Ω.
When switch S is closed as shown, the device functions at its maximum power of 5 W.
5.1 Explain, in words, the meaning of an emf of 12 V.
(2)
5.2 Calculate the current that passes through the
electrical device. (3)
5.3 Calculate the resistance of resistor Rx. (7)
5.4 Switch S is now opened. Will the device still
function at maximum power? Write down YES or
NO. Explain the answer without doing any
calculations. (4)
[16]
S
emf = 12 V
1 Ω
4 Ω
9 Ω
R
x
5 Ω
Electrical
device
3 Ω
A
C
B
D
QUESTION 5
A learner wants to use a 12 V battery with an internal resistance of 1 Ω to operate an electrical device. He
uses the circuit below to obtain the desired potential difference for the device to function. The resistance of
the device is 5 Ω.
When switch S is closed as shown, the device functions at its maximum power of 5 W.
5.1 Explain, in words, the meaning of an emf of 12 V.
(2)
5.2 Calculate the current that passes through the
electrical device. (3)
5.3 Calculate the resistance of resistor Rx. (7)
5.4 Switch S is now opened. Will the device still
function at maximum power? Write down YES or
NO. Explain the answer without doing any
calculations. (4)
[16]
S
emf = 12 V
1 Ω
4 Ω
9 Ω
R
x
5 Ω
Electrical
device
3 Ω
A
C
B
D
Grade 11 Electric circuits R Toerien Page 38
QUESTION 6
A battery of an unknown emf and an internal resistance of 0,5 Ω is connected to three resistors, a high-
resistance voltmeter and an ammeter of negligible resistance, as shown below.
The reading on the ammeter is 0,2 A.
6.1 Define the term potential difference. (2)
6.2 Calculate the reading on the voltmeter. (3)
6.3 Calculate the total current supplied by the battery. (4)
6.4 Calculate the emf of the battery. (5)
6.5 How would the voltmeter reading change if the 2 Ω
resistor is removed from the circuit? Write down
INCREASE, DECREASE , or REMAIN THE SAME.
Explain the answer. (4)
[18]
4 Ω 8 Ω
2 Ω
Ɛ
0,5Ω
V
A
QUESTION 6
A battery of an unknown emf and an internal resistance of 0,5 Ω is connected to three resistors, a high-
resistance voltmeter and an ammeter of negligible resistance, as shown below.
The reading on the ammeter is 0,2 A.
6.1 Define the term potential difference. (2)
6.2 Calculate the reading on the voltmeter. (3)
6.3 Calculate the total current supplied by the battery. (4)
6.4 Calculate the emf of the battery. (5)
6.5 How would the voltmeter reading change if the 2 Ω
resistor is removed from the circuit? Write down
INCREASE, DECREASE , or REMAIN THE SAME.
Explain the answer. (4)
[18]
4 Ω 8 Ω
2 Ω
Ɛ
0,5Ω
V
A
Grade 11 Electric circuits R Toerien Page 39
QUESTION 7
A battery with an internal resistance of 1 Ω and an unknown emf (Ɛ) is connected in a circuit, as shown
below. A high-resistance voltmeter V is connected across the battery. A1 and A2 represent ammeters of
negligible resistance.
With switch S closed, the current passing through the
8 Ω resistor is 0,5 A.
7.1 Define the term power in words. (2)
7.2 Calculate the reading on ammeter A1. (4)
7.3 If device R delivers power of 12 W, calculate the
reading on ammeter A2. (5)
7.4 Calculate the reading on the voltmeter when
switch S is open. (3)
[14]
V
S
Ɛ
1 Ω
R
20 Ω
A
2
8 Ω
16 Ω
A
1
QUESTION 7
A battery with an internal resistance of 1 Ω and an unknown emf (Ɛ) is connected in a circuit, as shown
below. A high-resistance voltmeter V is connected across the battery. A1 and A2 represent ammeters of
negligible resistance.
With switch S closed, the current passing through the
8 Ω resistor is 0,5 A.
7.1 Define the term power in words. (2)
7.2 Calculate the reading on ammeter A1. (4)
7.3 If device R delivers power of 12 W, calculate the
reading on ammeter A2. (5)
7.4 Calculate the reading on the voltmeter when
switch S is open. (3)
[14]
V
S
Ɛ
1 Ω
R
20 Ω
A
2
8 Ω
16 Ω
A
1
Grade 11 Electric circuits R Toerien Page 40
QUESTION 8
The headlamp and two IDENTICAL tail lamps of a scooter are connected in parallel to a battery with
unknown internal resistance as shown in the simplified circuit diagram. The headlamp has a resistance of
2,4 Ω and is controlled by switch S1. The tail lamps are controlled by switch S2. The resistance of the
connecting wires may be ignored.
The graph alongside shows the potential difference across the
terminals of the battery before and after switch S1 is closed (whilst
switch S2 is open). Switch S1 is closed at time t1.
8.1 Define the term current strength. (2)
8.2 Use the graph to determine the emf of the battery. (1)
8.3 With only switch S1 closed, calculate the following:
8.3.1 Current through the headlamp (3)
8.3.2 Internal resistance, r, of the battery (3)
8.4 Both switches S1 and S2 are now closed. The battery delivers
a current of 6 A during this period. Calculate the resistance of
each tail lamp. (5)
8.5 How will the reading on the voltmeter be affected if the
headlamp burns out? (Both switches S1 and S2 are still closed.)
Write down only INCREASES, DECREASES, or REMAINS
THE SAME. Explain the answer. (3)
[17]
S
1
2,4 Ω
Tail lamp 1
V
r
Tail lamp 2
S
2
Potential
difference (V)
Time (s)
12
t
1
9,6
QUESTION 8
The headlamp and two IDENTICAL tail lamps of a scooter are connected in parallel to a battery with
unknown internal resistance as shown in the simplified circuit diagram. The headlamp has a resistance of
2,4 Ω and is controlled by switch S1. The tail lamps are controlled by switch S2. The resistance of the
connecting wires may be ignored.
The graph alongside shows the potential difference across the
terminals of the battery before and after switch S1 is closed (whilst
switch S2 is open). Switch S1 is closed at time t1.
8.1 Define the term current strength. (2)
8.2 Use the graph to determine the emf of the battery. (1)
8.3 With only switch S1 closed, calculate the following:
8.3.1 Current through the headlamp (3)
8.3.2 Internal resistance, r, of the battery (3)
8.4 Both switches S1 and S2 are now closed. The battery delivers
a current of 6 A during this period. Calculate the resistance of
each tail lamp. (5)
8.5 How will the reading on the voltmeter be affected if the
headlamp burns out? (Both switches S1 and S2 are still closed.)
Write down only INCREASES, DECREASES, or REMAINS
THE SAME. Explain the answer. (3)
[17]
S
1
2,4 Ω
Tail lamp 1
V
r
Tail lamp 2
S
2
Potential
difference (V)
Time (s)
12
t
1
9,6
Grade 11 Electric circuits R Toerien Page 41
QUESTION 9
The circuit diagram below represents a combination of resistors in series and parallel. The battery has an
emf of 12 V and an unknown internal resistance r.
With switch S open, ammeter A gives a reading of 1,2 A.
9.1 Define the term resistance. (2)
9.2 Calculate the total resistance of the external circuit.
(2)
9.3 Calculate the internal resistance of the battery. (4)
9.4 Calculate the energy transferred in the 6 Ω resistor
in 3 minutes. (3)
Switch S is now closed.
9.5 How will each of the following be affected? Write
down only INCREASES, DECREASES , or REMAINS
THE SAME.
9.5.1 The total resistance of the circuit (1)
9.5.2 The reading on ammeter A (1)
9.6 A conducting wire of negligible resistance is now connected between points P and Q. What effect
will this have on the temperature of the battery? Write down only INCREASES, DECREASES, or
REMAINS THE SAME. Explain how you arrived at the answer. (4)
[17]
Q
P
6 Ω 3,6 Ω
emf = 12 V
r
8 Ω
S
A
QUESTION 9
The circuit diagram below represents a combination of resistors in series and parallel. The battery has an
emf of 12 V and an unknown internal resistance r.
With switch S open, ammeter A gives a reading of 1,2 A.
9.1 Define the term resistance. (2)
9.2 Calculate the total resistance of the external circuit.
(2)
9.3 Calculate the internal resistance of the battery. (4)
9.4 Calculate the energy transferred in the 6 Ω resistor
in 3 minutes. (3)
Switch S is now closed.
9.5 How will each of the following be affected? Write
down only INCREASES, DECREASES , or REMAINS
THE SAME.
9.5.1 The total resistance of the circuit (1)
9.5.2 The reading on ammeter A (1)
9.6 A conducting wire of negligible resistance is now connected between points P and Q. What effect
will this have on the temperature of the battery? Write down only INCREASES, DECREASES, or
REMAINS THE SAME. Explain how you arrived at the answer. (4)
[17]
Q
P
6 Ω 3,6 Ω
emf = 12 V
r
8 Ω
S
A
Grade 11 Electric circuits R Toerien Page 42
QUESTION 10
Learners want to construct an electric toaster
using one of two wires, A and B, made from
different materials. They conduct experiments
and draw the graphs as shown below.
10.1 Apart from temperature, write down
TWO other factors that the learners
should consider to ensure a fair test
when choosing which wire to use. (2)
10.2 Assuming all other factors are kept
constant, state which ONE of the two
wires will be the most suitable to use in
the toaster. Use suitable calculations to
show clearly how you arrive at the
answer. (8)
[10]
8,0
6,0
4,0
2,0
0,2 0,4 0,6 0,8 1,0
Current strength (A)
Potential difference (V)
0,0
Graph of V versus I for resistors A and B
A
B
QUESTION 10
Learners want to construct an electric toaster
using one of two wires, A and B, made from
different materials. They conduct experiments
and draw the graphs as shown below.
10.1 Apart from temperature, write down
TWO other factors that the learners
should consider to ensure a fair test
when choosing which wire to use. (2)
10.2 Assuming all other factors are kept
constant, state which ONE of the two
wires will be the most suitable to use in
the toaster. Use suitable calculations to
show clearly how you arrive at the
answer. (8)
[10]
8,0
6,0
4,0
2,0
0,2 0,4 0,6 0,8 1,0
Current strength (A)
Potential difference (V)
0,0
Graph of V versus I for resistors A and B
A
B
Grade 11 Electric circuits R Toerien Page 43
QUESTION 11
Learners conduct an investigation to determine the emf and internal
resistance (r) of a battery. They set up a circuit as shown in the circuit
diagram and measure the potential difference using the voltmeter for
different currents in the circuit.
The results obtained are shown in the graph.
11.1 Use the graph to determine the emf of
the battery. (1)
11.2 Calculate the gradient of the graph. (4)
11.3 Which physical quantity is represented
by the magnitude of the gradient of the
graph? (1)
11.4 How does the voltmeter reading change
as the ammeter reading increases? Write
down INCREASES, DECREASES , or
REMAINS THE SAME. Use the formula
emf = IR + Ir to explain the answer. (4)
[10]
A
V
r
Potential difference (V)
Current strength (A)
Graph of potential difference versus current
1,5
1,0
0,5
0
0,5 1,0 0,75 0,25
QUESTION 11
Learners conduct an investigation to determine the emf and internal
resistance (r) of a battery. They set up a circuit as shown in the circuit
diagram and measure the potential difference using the voltmeter for
different currents in the circuit.
The results obtained are shown in the graph.
11.1 Use the graph to determine the emf of
the battery. (1)
11.2 Calculate the gradient of the graph. (4)
11.3 Which physical quantity is represented
by the magnitude of the gradient of the
graph? (1)
11.4 How does the voltmeter reading change
as the ammeter reading increases? Write
down INCREASES, DECREASES , or
REMAINS THE SAME. Use the formula
emf = IR + Ir to explain the answer. (4)
[10]
A
V
r
Potential difference (V)
Current strength (A)
Graph of potential difference versus current
1,5
1,0
0,5
0
0,5 1,0 0,75 0,25
Grade 11 Electric circuits R Toerien Page 44
QUESTION 12 (DBE, Nov 2017)
The circuit below consists of three resistors, M, N, and T, a battery with emf Ɛ and an internal resistance of
0,9 Ω. The effective resistance between points a and b in the circuit is 6 Ω. The resistance of resistor T is
1,5 Ω.
When switch S is closed, a high-resistance voltmeter, V1, across a and b, reads 5 V.
Calculate the:
12.1 Current delivered by the battery (3)
12.2 Emf (Ɛ) of the battery (4)
Voltmeter V2 reads 2,5 V when the switch is closed.
12.3 Write down the resistance of N. (No calculations
required.) Give a reason for the answer. (2)
[9]
1,5 Ω
V
1
V
2
T
M N
S
a b
0,9
Ω
Ɛ
QUESTION 12 (DBE, Nov 2017)
The circuit below consists of three resistors, M, N, and T, a battery with emf Ɛ and an internal resistance of
0,9 Ω. The effective resistance between points a and b in the circuit is 6 Ω. The resistance of resistor T is
1,5 Ω.
When switch S is closed, a high-resistance voltmeter, V1, across a and b, reads 5 V.
Calculate the:
12.1 Current delivered by the battery (3)
12.2 Emf (Ɛ) of the battery (4)
Voltmeter V2 reads 2,5 V when the switch is closed.
12.3 Write down the resistance of N. (No calculations
required.) Give a reason for the answer. (2)
[9]
1,5 Ω
V
1
V
2
T
M N
S
a b
0,9
Ω
Ɛ
Grade 11 Electric circuits R Toerien Page 45
QUESTION 13 (DBE, June 2017)
The emf and internal resistance of a certain battery were determined experimentally. The circuit used for
the experiment is shown in the diagram. The data obtained from the experiment is presented in the table
below.
Current (A) Potential difference (V)
0,0 5,5
0,6 4,6
1,0 4,3
1,7 3,5
2,6 2,4
3,4 1,6
3,8 1,0
13.1 State Ohm's law in words. (2)
13.2 Plot a graph of V vs I and draw the line of
best fit through the plotted points. Ensure
that the line cuts both axes. (6)
Use the information in the graph to answer
QUESTIONS 13.3 and 13.4.
13.3 Write down the value of the emf (Ɛ) of the
battery. (1)
13.4 Determine the internal resistance of the
battery. (4)
[13]
TOTAL = 176
A
V
r
Ɛ
R
QUESTION 13 (DBE, June 2017)
The emf and internal resistance of a certain battery were determined experimentally. The circuit used for
the experiment is shown in the diagram. The data obtained from the experiment is presented in the table
below.
Current (A) Potential difference (V)
0,0 5,5
0,6 4,6
1,0 4,3
1,7 3,5
2,6 2,4
3,4 1,6
3,8 1,0
13.1 State Ohm's law in words. (2)
13.2 Plot a graph of V vs I and draw the line of
best fit through the plotted points. Ensure
that the line cuts both axes. (6)
Use the information in the graph to answer
QUESTIONS 13.3 and 13.4.
13.3 Write down the value of the emf (Ɛ) of the
battery. (1)
13.4 Determine the internal resistance of the
battery. (4)
[13]
TOTAL = 176
A
V
r
Ɛ
R
Grade 11 Electric circuits R Toerien Page 46
Grade 11 Electric circuits R Toerien Page 47
Grade 11 Physics Module P5 CAPS Electric circuits
Ohm’s Law
• Determine the relationship between current, voltage and resistance at a constant temperature using
a simple circuit
• State the difference between Ohmic and non-Ohmic conductors, and give an example of each
• Solve problems using the mathematical expression of Ohm’s Law, R =
V
I
, for series and parallel
circuits
Internal resistance and series-and parallel networks
• Solve problems involving current, voltage, and resistance for circuits containing arrangements of
resistors in series and in parallel
• State that a real battery has internal resistance
• The sum of the voltages across the external circuit plus the voltage across the internal resistance is
equal to the emf: Ɛ = Vload + Vinternal resistance or Ɛ = IRext + Ir
• Solve circuit problems in which the internal resistance of the battery must be considered.
• Solve circuit problems, with internal resistance, involving series-parallel networks of resistors
Power and Energy
• Define power as the rate at which electrical energy is converted in an electric circuit and is
measured in watts (W)
• Know that electrical power dissipated in a device is equal to the product of the potential difference
across the device and current flowing through it i.e. P = VI
• Know that power can also be given by P = I
2
R or P =
V
2
R
• Solve circuit problems involving the concept of power
• Know that the electrical energy is given by E = PΔt and is measured in joules (J)
• Solve problems involving the concept of electrical energy
• Know that the kilowatt-hour (kWh) refers to the use of 1 kilowatt of electricity for 1 hour
• Calculate the cost of electricity usage given the power specifications of the appliances used, as well
as the duration, if the cost of 1 kWh is given
Grade 11 Physics Module P5 CAPS Electric circuits
Ohm’s Law
• Determine the relationship between current, voltage and resistance at a constant temperature using
a simple circuit
• State the difference between Ohmic and non-Ohmic conductors, and give an example of each
• Solve problems using the mathematical expression of Ohm’s Law, R =
V
I
, for series and parallel
circuits
Internal resistance and series-and parallel networks
• Solve problems involving current, voltage, and resistance for circuits containing arrangements of
resistors in series and in parallel
• State that a real battery has internal resistance
• The sum of the voltages across the external circuit plus the voltage across the internal resistance is
equal to the emf: Ɛ = Vload + Vinternal resistance or Ɛ = IRext + Ir
• Solve circuit problems in which the internal resistance of the battery must be considered.
• Solve circuit problems, with internal resistance, involving series-parallel networks of resistors
Power and Energy
• Define power as the rate at which electrical energy is converted in an electric circuit and is
measured in watts (W)
• Know that electrical power dissipated in a device is equal to the product of the potential difference
across the device and current flowing through it i.e. P = VI
• Know that power can also be given by P = I
2
R or P =
V
2
R
• Solve circuit problems involving the concept of power
• Know that the electrical energy is given by E = PΔt and is measured in joules (J)
• Solve problems involving the concept of electrical energy
• Know that the kilowatt-hour (kWh) refers to the use of 1 kilowatt of electricity for 1 hour
• Calculate the cost of electricity usage given the power specifications of the appliances used, as well
as the duration, if the cost of 1 kWh is given
Grade 11 Electric circuits R Toerien Page 48
Grade 11 Physics Module P5 Definitions Electric circuits
Potential difference: It is the energy transferred per unit charge flowing between two points
through a conductor.
Emf: It is the energy provided by a battery per unit charge flowing through the
battery.
Terminal potential difference: The voltage measured across the terminals of a battery when charges are
flowing in the circuit.
Current strength: The rate of flow of charge.
Resistance: The ratio of the potential difference across a resistor to the current in the
resistor.
Internal resistance: The resistance within a battery when current is flowing which causes a
decrease in the potential difference that the battery can supply to the
external circuit.
One ohm: One ohm is equal to one volt per ampere.
One coulomb: The charge transferred in a conductor in one second if the current is one
ampere.
Ohm's law: The potential difference across a conductor is directly proportional to the
current in the conductor at constant temperature.
Power: It is the rate at which work is done.
Grade 11 Physics Module P5 Definitions Electric circuits
Potential difference: It is the energy transferred per unit charge flowing between two points
through a conductor.
Emf: It is the energy provided by a battery per unit charge flowing through the
battery.
Terminal potential difference: The voltage measured across the terminals of a battery when charges are
flowing in the circuit.
Current strength: The rate of flow of charge.
Resistance: The ratio of the potential difference across a resistor to the current in the
resistor.
Internal resistance: The resistance within a battery when current is flowing which causes a
decrease in the potential difference that the battery can supply to the
external circuit.
One ohm: One ohm is equal to one volt per ampere.
One coulomb: The charge transferred in a conductor in one second if the current is one
ampere.
Ohm's law: The potential difference across a conductor is directly proportional to the
current in the conductor at constant temperature.
Power: It is the rate at which work is done.
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