ch_21_PPT_lecture for physics resitor in series and parallel circuit

ch_21_PPT_lecture for physics resitor in series and parallel circuit

1. Bell Ringer Activity Think-Pair-Share Everyday Physics in your life: Imagine you have a circuit with a light bulb connected to a battery. Describe in your own words what you think happens to the flow of electric charge and the potential difference as the circuit is completed by closing the switch . How do you think the current and potential difference relate to each other in this scenario? © 2014 Pearson Education,Inc .

2. Learning Outcomes Solve problems related to equivalent resistance. Design an experiment to verify Ohm's Law. Apply your understanding of resistance to real life.

3. SUCCESS CRITERIA I can define resistance. I can state Ohm’s Law. I can interpret experimental data to determine the relationship between voltage, current, and resistance. I can relate the concept of resistance to real life (UAE)

4. Oral presentation : Resistors and resistance Write From Scratch Using Grammarly's Generative AI (youtube.com) Questions /Self assess.

5. Ohm’s Law Ohms law V I R - Google Search To move electrons against the resistance of a wire, it is necessary to apply a potential difference between the wire ' s ends. Ohm ' s law relates the applied potential difference to the current produced and the wire ' s resistance . To be specific, the three quantities are related as follows: © 2014 Pearson Education, Inc.

6. Ohm’s Law Ohm ' s law is named for the German physicist Georg Simon Ohm (1789–1854). Rearranging Ohm ' s law to solve for the resistance, we find R = V / I From this expression, it is clear that resistance has units of volts per amp. A resistance of 1 volt per amp defines a new unit—the ohm . The Greek letter omega ( Ω ) is used to designate the ohm. Thus, 1 Ω = 1 V/A A device for measuring resistance is called an ohmmeter. © 2014 Pearson Education, Inc.

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8. I –We –YOU strategy A potential difference of 24 V is applied to a 150 Ω resistor. How much current flows through the resistor ? Given Equation Potential difference= 24 V V = R x I R =150 Ω I = V /R I = ?? I = 24 /150 = 0.16 A © 2014 Pearson Education, Inc.

9. WE – Group work A potential difference of 16 V is applied to a resistor whose resistance is 220 Ω .What is the current that flows through the resistor? © 2014 Pearson Education, Inc.

10. YOU – Check your channel on teams! Question 1: gas group What voltage is required to produce a current of 1.8 A through a 140 Ω resistor? Question 2:Liquid group When a potential difference of 18 V is applied to a given wire ,the wire conducts 0.35 A of current. What is the resistance of the wire? Question 3:Solid group A current of 0.2A passes through a 1.4 kΩ  resistor. What is the voltage across it? © 2014 Pearson Education, Inc.

11. Bell Ringer Activity Think-Pair-Share Use Padlet the answer the following questions: Everyday Physics in your life: How does the function of a resistor differ from other components in a circuit? Describe the relationship between voltage, current, and resistance using Ohm's Law.

12. Series and Parallel Circuits | Electricity | Physics | FuseSchool (youtube.com)

13. Check Point ( Self-Assessment ) 1.In a parallel circuit, what is true about the potential difference across each component? a. It is different for each component b. It is equal for each component c. It varies based on resistance 2.True/False Total voltage in series circuits is not shared between components. 3. Explain how does the current vary in series circuits.

14. EMSAT ‘STYLE’ question : Pair work

15. Electric Circuits Electric circuits often contain a number of resistors connected in various ways. One way resistors can be connected is end to end. Resistors connected in this way are said to form a series circuit. The figure below shows three resistors R 1 , R 2 , and R 3 , connected in series. © 2014 Pearson Education, Inc.

16. Electric Circuits The three resistors acting together have the same effect—that is, they draw the same current—as a single resistor, which is referred to as the equivalent resistor, R eq . This equivalence is illustrated in the figure below. The equivalent resistor has the same current, I , flowing through it as each resistor in the original circuit. © 2014 Pearson Education, Inc.

17. Electric Circuits When resistors are connected in series, the equivalent resistance is simply the sum of the individual resistances. In our case, with three resistors, we have R eq = R 1 + R 2 + R 3 In general, the equivalent resistance of resistors in series is the sum of all the resistances that are connected together: © 2014 Pearson Education, Inc.

18. Electric Circuits The following example illustrates the functioning of a series circuit. © 2014 Pearson Education, Inc.

19. Electric Circuits © 2014 Pearson Education, Inc.

20. Electric Circuits Resistors that are connected across the same potential difference are said to form a parallel circuit. An example of three resistors connected in parallel is shown the figure below. © 2014 Pearson Education, Inc.

21. Electric Circuits In a case like this, the electrons have three parallel paths through which they can flow—like parallel lanes on the highway. The three resistors acting together draw the same current as a single equivalent resistor, R eq , as indicated in the figure below. © 2014 Pearson Education, Inc.

22. Electric Circuits When resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances. Thus, for our circuit of three resistors, we have 1/ R eq = 1/ R 1 + 1/ R 2 + 1/ R 3 In general, the inverse equivalent resistance is equal to the sum of all of the individual inverse resistances: © 2014 Pearson Education, Inc.

23. Electric Circuits As an example of parallel resistors, consider a circuit with two identical resistors, R , connected in parallel. The equivalent resistance in this case is 1/ R eq = 1/ R + 1/ R 1/ R eq = 2/ R © 2014 Pearson Education, Inc.

24. WE – Group work © 2014 Pearson Education, Inc. Calculate the total resistance of the following resistors. What is the equivalent resistance if 3Ω, 20Ω and 32Ω are connected in series. What is the equivalent resistance if 34 Ω and 20 Ω are connected in parallel.

25. You –Check your Channel on Teams! Calculate the total resistance in each of the following circuit © 2014 Pearson Education, Inc. Gas group Liquid group Solid group

26. Electric Circuits The following example illustrates the functioning of a parallel circuit. © 2014 Pearson Education, Inc.

27. Electric Circuits

28. Electric Circuits By considering the resistors in pairs or groups that are connected in parallel or in series, you can reduce the entire circuit to one equivalent circuit. This method is applied in the following example. © 2014 Pearson Education, Inc.

29. Electric Circuits

30. Electric Circuits A voltmeter is a device used to measure the potential difference between any two points in a circuit. To measure the voltage between two points, for example, points C and D in the figure below, the voltmeter is placed in parallel at the appropriate points. © 2014 Pearson Education, Inc.

31. Power and Energy in Electric Circuits The power delivered by an electric circuit increases with both the current and the voltage. Increase either, and the power increases. When a ball falls in a gravitational field, there is a change in gravitational potential energy. Similarly, when an amount of charge, Δ Q , moves across a potential difference, V , there is a change in electrical potential energy, Δ PE , given by Δ PE = ( Δ Q ) V © 2014 Pearson Education, Inc.

32. Power and Energy in Electric Circuits Recalling that power is the rate at which energy changes, P = Δ E / Δ t , we can express the electric power as follows: P = Δ E / Δ t = ( Δ Q ) V / Δ t Knowing that the electric current is given by I = ( Δ Q )/ Δ t allows us to write an expression for the electric power in terms of the current and voltage. © 2014 Pearson Education, Inc.

33. Power and Energy in Electric Circuits Thus, the electric power used by a device is equal to the current times the voltage. For example, a current of 1 amp flowing across a potential difference of 1 V produces a power of 1 W. The following example provides another example of how the electric power is calculated. © 2014 Pearson Education, Inc.

34. Power and Energy in Electric Circuits The equation P = IV applies to any electrical system. In the special case of a resistor, the electric power is dissipated in the form of heat and light, as shown in the figure, where the electric power dissipated in an electric space heater. © 2014 Pearson Education, Inc.

35. Power and Energy in Electric Circuits Applying Ohm ' s law, V = IR , which deals with resistors, we can express the power dissipated in a resistor as follows: P = IV = I ( IR ) = I 2 R Similarly, solving Ohm ' s law for the current, I = V / R , and substituting that result gives an alternative expression for the power dissipated in a resistor: P = IV = ( V / R ) V = V 2 / R All three equations for power are valid. The first, P = IV , applies to all electrical systems. The other two ( P = I 2 R and P = V 2 / R ) are specific to resistors, which is why the resistance, R , appears in those equations. © 2014 Pearson Education, Inc.

36. Power and Energy in Electric Circuits The filament of an incandescent lightbulb is basically a resistor inside a sealed, evacuated tube. The filament gets so hot that it glows, just like the heating coil on a stove or the coils in a space heater. The power dissipated in the filament determines the brightness of the lightbulb . The higher the power, the brighter the bulb. This basic concept is applied in the example on the next slide. © 2014 Pearson Education, Inc.

37. Power and Energy in Electric Circuits © 2014 Pearson Education, Inc.

38. Power and Energy in Electric Circuits The local electric company bills consumers for the electricity they use each month. To do this, they use a convenient unit for measuring electric energy called the kilowatt-hour. Recall that a kilowatt is 1000 W, or equivalently, 1000 J/s. Similarly, an hour is 3600 s. Combining these results, we see that a kilowatt-hour is equal to 3.6 million joules of energy: 1 kWh = (1000 J/s)(3600 s) = 3.6 x 10 6 J © 2014 Pearson Education, Inc.

39. Power and Energy in Electric Circuits The figure below shows the type of meter used to measure the electrical energy consumption of a household, as well as the typical bill. © 2014 Pearson Education, Inc.

40. Power and Energy in Electric Circuits The following example illustrates how the cost of electrical energy is calculated. © 2014 Pearson Education, Inc.

41. Power and Energy in Electric Circuits