Physics 102 Lecture 06 Exam I Kirchhoffs Laws

Physics 102: Lecture 06 Exam I Kirchhoff’s Laws PRACTICE EXAM IS AVAILABLE-CLICK ON “COURSE INFO” ON WEB PAGE Conflict Exam sign up available in grade book Be Careful with round off errors in homework 3!

Last Time • Resistors in series: Last Lecture Current thru is same; Voltage drop across is IRi • Resistors Voltage drop across is same; Current thru is V/Ri • Solved Today in parallel: • What Circuits about this one? 5

Good News and Bad News Good News: You already know everything you need to get 100% on circuit problems for Hour Exam 1! Bad News: You don’t realize it! So…. Let’s see what you know! 10

Kirchhoff’s Rules • Kirchhoff’s Junction Rule (KJR): – Current going in equals current coming out. • Kirchhoff’s Loop Rule (KLR): – Sum of voltage drops around a loop is zero. 12

Using Kirchhoff’s Rules (1) Label all currents Choose any direction (2) Label +/- for all elements R 1 A + I 1 - + Current goes + - (for resistors) (3) Choose loop and direction + B E 1 - E 2 (4) Write down voltage drops 1. Be careful about signs 1. + R 2 E 3 I 2 I 3 - + R 3 R 5 - + I 4 - R 4 + + 17

Loop Rule Practice R 1=5 W Find I: I B e 1= 50 V A R 2=15 W e 2= 10 V 22

Loop Rule Practice R 1=5 W Find I: Label currents Label elements +/Choose loop Write KLR B I - + + e 1= 50 V - A - + R 2=15 W - + e 2= 10 V –e 1+IR 1 + e 2 + IR 2 = 0 -50 + 5 I + 10 +15 I = 0 I = +2 Amps 22

ACT: KLR 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. R 1=10 W I 1 E 2 = 5 V I 2 R 2=10 W IB + E 1 = 10 V Upper loop contains R 1 and R 2 but also E 2. 25

Preflight 6. 1 Calculate the current through resistor 1. 28% 62% 1) I 1 = 0. 5 A 2) I 1 = 1. 0 A 10% 3) I 1 = 1. 5 A -E 1 + I 1 R = 0 I 1 = E 1 /R = 1 A I 1 R=10 W + - E 2 = 5 V I 2 R=10 W IB + E 1 = 10 V 27

Preflight 6. 1 Calculate the current through resistor 1. 20% 63% 1) I 1 = 0. 5 A 2) I 1 = 1. 0 A 17% 3) I 1 = 1. 5 A I 1 R=10 W + - E 2 = 5 V I 2 -E 1 + I 1 R = 0 I 1 = E 1 /R = 1 A R=10 W IB + E 1 = 10 V ACT: Voltage Law How would I 1 change if the switch was opened? 1) Increase 2) No change 3) Decrease 32

Preflight 6. 2 Calculate the current through resistor 2. 53% 31% 1) I 2 = 0. 5 A 2) I 2 = 1. 0 A 16% 3) I 2 = 1. 5 A R=10 W I 1 E 2 = 5 V I 2 -E 1 +E 2 + I 2 R = 0 I 2 = 0. 5 A R=10 W + IB + E 1 = 10 V 35

Preflight 6. 2 How do I know the direction of I 2? It doesn’t matter. Choose whatever direction you like. Then solve the equations to find I 2. If the result is positive, then your initial guess was correct. If result is negative, then actual direction is opposite to your initial guess. R=10 W I 1 E 2 = 5 V I 2 Work through preflight with opposite sign for I 2? R=10 W + - IB + E 1 = 10 V -E 1 +E 2 - I 2 R = 0 Note the sign change from last slide I 2 = -0. 5 A Answer has same magnitude as before but opposite sign. That means current goes to the left, as we found before. 35

Kirchhoff’s Junction Rule Current Entering = Current Leaving I 1 = I 2 + I 3 I 2 I 3 Preflight 6. 3 18% 37% R=10 W I 1 44% E=5 V I 2 R=10 W 1) IB = 0. 5 A 2) IB = 1. 0 A 3) IB = 1. 5 A IB = I 1 + I 2 = 1. 5 A “The first two can be calculated using V=IR because the voltage and resistance is given, and the current through E 1 can be calculated with the help of Kirchhoff's Junction rule, that states whatever current flows into the junction IB + E 1 = 10 V 38

Kirchhoff’s Laws (1) Label all currents Choose any direction (2) Label +/- for all elements R 1 I 1 A R 2 Current goes + - (for resistors) (3) Choose loop and direction Your choice! (4) Write down voltage drops (5) Follow any loops B E 1 E 3 I 2 I 3 R 3 E 2 I 4 R 5 (5) Write down junction equation (6) Iin = Iout 39 36

You try it! In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. + R 1 - I 1 I 3 I 2 e 1 + - R 2 - R 3 + - e 2 + + - 45

You try it! In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. 1. 2. 3. 4. Label all currents Label +/- for all elements (Choose any direction) (Current goes + - for resistor) Choose loop and direction (Your choice!) Write down voltage drops (First sign you hit is sign to use!) Loop 1: – e 1+I 1 R 1 – I 2 R 2 = 0 Loop 2: + I 2 R 2 + I 3 R 3 + e 2 = 0 5. + R 1 - I 1 I 3 I 2 Write down junction equation Node: I 1 + I 2 = I 3 + e 1 - 3 Equations, 3 unknowns the rest is math! Loop 1 R 2 + - R 3 + - Loop 2 - e 2 + 45

Let’s put in real numbers In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. + 5 - I 1 I 3 I 2 + 20 - 10 + - 10 - + - 2 1. left loop: -20+5 I 1 -10 I 2 = 0 2. outer loop: -20 +5 I 1+10 I 3+2=0 3. junction: I 3=I 1+I 2 + solution: substitute Eq. 3 for I 3 in Eq. 2: -20 + 5 I 1 + 10(I 1+I 2) + 2 = 0 rearrange: 15 I 1+10 I 2 = 18 rearrange Eq. 1: 5 I 1 -10 I 2 = 20 Now we have 2 eq. , 2 unknowns. Continue on next slide 45

15 I 1+10 I 2 = 18 5 I 1 - 10 I 2 = 20 Now we have 2 eq. , 2 unknowns. Add the equations together: 20 I 1=38 I 1=1. 90 A U′ Plug into bottom equation: 5(1. 90)-10 I 2 = 20 I 2=-1. 05 A note that this means direction of I 2 is opposite to that shown on the previous slide Use junction equation I 3=I 1+I 2 = 1. 90 -1. 05 I 3 = 0. 85 A We are done!

Some practice from the book: 18. 52, 18. 53, 18. 54, 18. 55, 18. 67 See you Wednesday