Kirchhoff’s laws

NCEA Level 3 Physics Electricity AS91526 Kirchhoff's Laws

1. Kirchhoff’s Laws What goes in must come out

2. Kirchhoff’s Current Law The current that flows into a junction must equal the current flowing out of a junction 5 A (in) 3 A (out) 2 A (out) That all makes perfect sense; you can’t get more current out that you put in

3. Kirchhoff’s Voltage Law The sum of the voltages around any circuit loop is zero (voltage produced must equal the voltage used in a loop) 14V (produced) 7V (used) 7V (used) Call this voltage produced positive +14V Call these voltages used negative -7V This makes sense too: you can’t be using more energy that you are producing

4. Getting more complex Some loops involve more than one cell or battery and several components. It is important to work your way around a loop systematically in one direction (doesn’t matter which way you go as long as you keep going in the same direction around the whole loop)

5. The Rules -Cells Passing through a cell in the same direction as conventional current is a positive voltage (energy produced) Passing through a cell in the opposite direction to conventional current is a negative voltage (energy used) 12V I 12V I +12V -12V

6. The Rules -Components Passing through a component in the same direction as conventional current is a negative voltage (energy used) Passing through a component in the opposite direction to conventional current is a positive voltage (energy gained) -12V +12V

7. Putting it all together Working anticlockwise around the loop ABCDA AB +15V BC –(2X11)V ( V=IR) CD +(4X3)V -5V so: +15 - 22 +12 – 5 = 0 2A 15V 5V 4 11 A B C D 5A

8. 5A 2A 15V 5V 4 11 A B C D Try the other direction Now clockwise around the loop BADCB BA -15V DC +5V –(4X3)V (V=IR) CB +(11X2)V so: -15 + 5 - 12 + 22 = 0

9. Try this; Working anticlockwise around the loop ABCDA AB +15V BC –(1.5X40)V (V=IR) CD +(2.5X20)V -5V so: +15 - 60 +50 – 5 = 0 4A 15V 5V 20 40 A B C D 2.5A

10. 7A 3A 5V 12V 4 ? B A D C What if something is missing? Now clockwise around the loop ABCDA AB -5V CD +12V –(4X4)V (V=IR) DA +(3X R )V so: -5 + 12 – 16 + 3 R = 0 3 R =9 R =3 

11. or this one… Now clockwise around the loop ABCDA AB -15V CD + V –(0.5X80)V (V=IR) DA +(1.5X30)V so: -15 + V – 40 + 45 = 0 V =10V 0.5A 15V ?V 80 30 B A D C 2A

12. Now some Exercises Try ESA, Activity 13C, Pg 214 ABA, Pg 134-136

13. Homework