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!

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

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.

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 I4I R1R1 E1E1 R2R2 R3R3 E2E2 E3E3 R5R5 A B

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

Physics 102: Lecture 6, Slide 6 KVR Practice R 1 =5  I –  1 +IR 1 +  2 + IR 2 = I 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  = -40 Volts V A - V B = –E 1 + IR 1 =  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

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 + -

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

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

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

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.

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.

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

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

Physics 102: Lecture 6, Slide 15 You try it! R1R1 R2R2 R3R3 I1I1 I3I3 I2I 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

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

Physics 102: Lecture 6, Slide 17 You try it! R1R1 R2R2 R3R3 I1I1 I3I3 I2I 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  

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