1. Physics 102: Lecture 6, Slide 1 Kirchhoff’s Laws Today’s lecture will cover Textbook Sections 18.5, 7 Physics 102: Lecture 06 Conflict Exam sign up available in grade book Be Careful with round off errors in homework 3!

2. Physics 102: Lecture 6, Slide 2 Last Time Resistors in series: Resistors in parallel: Current thru is same; Voltage drop across is IR i Voltage drop across is same; Current thru is V/R i Last Lecture Solved Circuits What about this one? Today

3. Physics 102: Lecture 6, Slide 3 Kirchhoff’s Rules Kirchhoff’s Voltage Rule (KVR):KVR –Sum of voltage drops around a loop is zero. Kirchhoff’s Current Rule (KCR): –Current going in equals current coming out.

4. Physics 102: Lecture 6, Slide 4 Kirchhoff’s Laws (1)Label all currents Choose any direction (2)Label +/- for all elements Current goes +  - (for resistors) (3)Choose loop and direction Must start on wire, not element. (4)Write down voltage drops First sign you hit is sign to use. R4R4 I1I1 I3I3 I2I2 I4I4 + + ++ + - -- - - + + + - - - R1R1 E1E1 R2R2 R3R3 E2E2 E3E3 R5R5 A B

5. Physics 102: Lecture 6, Slide 5 KVR Practice R 1 =5  I  1 = 50V + - + - + - +- R 2 =15   2 = 10V A B What if only went from A to B? V A – V B = V A - V B = Find I:  2 = 10V R 1 =5  I + - + -  1 = 50V + - +- R 2 =15  A B 22

6. Physics 102: Lecture 6, Slide 6 KVR Practice R 1 =5  I –  1 +IR 1 +  2 + IR 2 = 0 -50 + 5 I + 10 +15 I = 0 I = +2 Amps  1 = 50V + - + - + - +- R 2 =15   2 = 10V A B What if only went from A to B? V A - V B = -IR 2 – E 2 = -2  15 - 10 = -40 Volts V A - V B = –E 1 + IR 1 = -50 + 2  5 = -40 Volts Find I:  2 = 10V R 1 =5  I + - + -  1 = 50V + - +- R 2 =15  A B Label currents Label elements +/- Choose loop Write KVR

7. Physics 102: Lecture 6, Slide 7 ACT: KVR R 1 =10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R 2 =10  I2I2 Resistors R 1 and R 2 are: 1) in parallel 2) in series 3) neither + -

8. Physics 102: Lecture 6, Slide 8 ACT: KVR R 1 =10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R 2 =10  I2I2 Resistors R 1 and R 2 are 1) in parallel 2) in series 3) neither + - Definition of parallel: Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. Definition of series: Two elements are in series if (and only if) every loop that Contains R 1 also contains R 2

9. Physics 102: Lecture 6, Slide 9 Preflight 6.1 R 1 =10  E 1 = 10 V IBIB I1I1 R 2 =10  I2I2 1)I 1 = 0.5 A 2)I 1 = 1.0 A 3)I 1 = 1.5 A +- + - How would I 1 change if the switch was closed? E 2 = 5 V 1) Increase 2) No change 3) Decrease Calculate the current through resistor 1. ACT: Voltage Law

10. Physics 102: Lecture 6, Slide 10 Preflight 6.1 R 1 =10  E 1 = 10 V IBIB I1I1 R 2 =10  I2I2 1) I 1 = 0.5 A 2) I 1 = 1.0 A 3) I 1 = 1.5 A +- + - - E 1 + I 1 R 1 = 0  I 1 = E 1 /R 1 = 1A How would I 1 change if the switch was opened? E 2 = 5 V 1) Increase 2) No change 3) Decrease Calculate the current through resistor 1. ACT: Voltage Law

11. Physics 102: Lecture 6, Slide 11 Preflight 6.2 R 1 =10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R 2 =10  I2I2 1)I 2 = 0.5 A 2)I 2 = 1.0 A 3)I 2 = 1.5 A + - + - Calculate the current through resistor 2.

12. Physics 102: Lecture 6, Slide 12 Preflight 6.2 R 1 =10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R 2 =10  I2I2 1) I 2 = 0.5 A 2) I 2 = 1.0 A 3) I 2 = 1.5 A + - + - - E 1 + E 2 + I 2 R 2 = 0  I 2 = 0.5A Calculate the current through resistor 2.

13. Physics 102: Lecture 6, Slide 13 Kirchhoff’s Junction Rule Current Entering = Current Leaving I1I1 I2I2 I3I3 I 1 = I 2 + I 3 1)I B = 0.5 A 2)I B = 1.0 A 3)I B = 1.5 A R=10  E 1 = 10 V IBIB I1I1 E = 5 V R=10  I2I2 + - Preflight 6.3

14. Physics 102: Lecture 6, Slide 14 Kirchhoff’s Junction Rule Current Entering = Current Leaving I1I1 I2I2 I3I3 I 1 = I 2 + I 3 1) I B = 0.5 A 2) I B = 1.0 A 3) I B = 1.5 A I B = I 1 + I 2 = 1.5 A R=10  E 1 = 10 V IBIB I1I1 E = 5 V R=10  I2I2 + - Preflight 6.3

15. Physics 102: Lecture 6, Slide 15 You try it! R1R1 R2R2 R3R3 I1I1 I3I3 I2I2 + - + + + Loop 1: 1.Label all currents 2. Label +/- for all elements 3. Choose loop and direction 4.Write down voltage drops - - - Loop 2: 11 5. Write down node equation Node: 22 In the circuit below you are given  1,  2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. + -

16. Physics 102: Lecture 6, Slide 16 Kirchhoff’s Laws (1)Label all currents Choose any direction (2)Label +/- for all elements Current goes +  - (for resistors) (3)Choose loop and direction Your choice! (4)Write down voltage drops Follow any loops (5)Write down node equation I in = I out R4R4 R1R1 E1E1 R2R2 R3R3 E2E2 E3E3 I1I1 I3I3 I2I2 I4I4 R5R5 A B 36

17. Physics 102: Lecture 6, Slide 17 You try it! R1R1 R2R2 R3R3 I1I1 I3I3 I2I2 + - + + + Loop 1: –  1 +I 1 R 1 – I 2 R 2 = 0 1.Label all currents (Choose any direction) 2. Label +/- for all elements (Current goes +  - for resistor) 3. Choose loop and direction (Your choice!) 4.Write down voltage drops (First sign you hit is sign to use!) - - - Loop 2: 11 5. Write down node equation Node: I 1 + I 2 = I 3 22 3 Equations, 3 unknowns the rest is math! In the circuit below you are given  1,  2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. Loop 1 Loop 2  + -   + I 2 R 2 + I 3 R 3 +  2 = 0  

18. Physics 102: Lecture 6, Slide 18 See you next lecture! Read Sections 18.10,11