Kirchhoffs Laws Dorsey DE PHYS 202 Kirchhoffs Rules

Kirchhoff’s Laws Dorsey, DE PHYS 202

Kirchhoff’s Rules • Kirchhoff’s Junction Rule: – Current going in equals current coming out. • Kirchhoff’s Loop Rule: – Sum of voltage change around a loop is zero.

Using Kirchhoff’s Rules (1) Label all currents (2) Write down junction equation Iin = Iout (3)Choose loop and direction • • Choose any direction You will need one less loop than unknown currents (4) Write down voltage changes Be careful about signs • For batteries – voltage change is positive when going from negative to positive • For resistors – voltage change is negative when going the direction of the current R 1 I 1 A R 2 B 1 I 2 3 I 3 R 3 2 I 5 R 5 I 4 R 4

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

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

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

Loop Rule Practice R 1=5 W Find I: Label currents Choose loop Write KLR I B 1= 50 V A +e 1 - IR 1 - e 2 - IR 2 = 0 +50 - 5 I - 10 - 15 I = 0 +40 – 20 I = +2 Amps R 2=15 W 2= 10 V

Resistors R 1 and R 2 are 1. In parallel 2. In series 3. neither R 1=10 W I 1 E 2 = 5 V I 2 R 2=10 W IB + E 1 = 10 V

Resistors R 1 and R 2 are 1. In parallel 2. In series 3. neither R 1=10 W I 1 E 2 = 5 V I 2 R 2=10 W IB Definition of parallel: + E 1 = 10 V Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. Upper loop contains R 1 and R 2 but also E 2.

Preflight 10. 1 Calculate the current through resistor 1. 1) I 1 = 0. 5 A 2) I 1 = 1. 0 A 3) I 1 = 1. 5 A R=10 W I 1 E 2 = 5 V I 2 R=10 W IB E 1 = 10 V 27

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

How would I 1 change if the switch was opened? R=10 W I 1 1. Increase 2. No change 3. Decrease E 2 = 5 V I 2 R=10 W IB E 1 = 10 V

How would I 1 change if the switch was opened? R=10 W I 1 1. Increase 2. No change 3. Decrease E 2 = 5 V I 2 R=10 W IB E 1 = 10 V

Preflight 10. 2 Calculate the current through resistor 2. 1) I 2 = 0. 5 A 2) I 2 = 1. 0 A R=10 W I 1 E 2 = 5 V I 2 R=10 W 3) I 2 = 1. 5 A IB E 1 = 10 V 35

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

Preflight 10. 2 How do I know the direction to draw I 2? It doesn’t matter! Choose whatever, 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 guess. Work through this example with opposite sign for I 2? +E 1 - E 2 + I 2 R = 0 Note the sign change from last slide R=10 W I 1 E 2 = 5 V I 2 R=10 W + - IB + E 1 = 10 V

Preflight 10. 2 How do I know the direction to draw I 2? It doesn’t matter! Choose whatever, 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 guess. Work through this example with opposite sign for I 2? +E 1 - E 2 + I 2 R = 0 Note the sign change from last slide 10 – 5 + 10 I = 0 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. R=10 W I 1 E 2 = 5 V I 2 R=10 W + - IB + E 1 = 10 V

Kirchhoff’s Junction Rule Find the total current flowing through the point Current Entering = Current Leaving I 3 = I 1 + I 2 I 1 I 2 I 3 R=10 W I 1 1) IB = 0. 5 A 2) IB = 1. 0 A 3) IB = 1. 5 A E=5 V I 2 R=10 W IB + E 1 = 10 V

Kirchhoff’s Junction Rule Find the total current flowing through the point Current Entering = Current Leaving I 3 = I 1 + I 2 I 1 I 2 I 3 10 – 10 I 1 = 0 so, I 1 = 1 10 – 5 – 10 I 2 = 0 so, I 2 =. 5 1) IB = 0. 5 A 2) IB = 1. 0 A 3) IB = 1. 5 A R=10 W I 1 E=5 V I 2 R=10 W I 3 = 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 must flow out. So I 1 and I 2 are added together. ” I 3 + E 1 = 10 V

Kirchhoff’s Laws (Review) (1) Label all currents Choose any direction (2) Write down the junction equation R 1 I 1 A Iin = Iout (3) Choose loop and direction Your choice! (4) Write down voltage changes Follow any loops (5) Solve the equations by substitution or combination. R 2 B E 1 E 3 I 2 I 3 R 3 E 2 R 5 I 4 R 4

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

You try it! In the circuit below you are given 1, 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. 1. 2. 3. Label all currents (Choose any direction) Write down junction equation Node: I 1 + I 2 = I 3 Choose loop and direction (Your choice!) 4. Write down voltage changes 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 3 Equations, 3 unknowns the rest is math! R 1 I 3 I 1 I 2 1 + - Loop 1 R 2 R 3 Loop 2 - + 2

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

-10 I 1 -20 I 2 = 2 2*(5 I 1 - 10 I 2 = 20) = 10 I 1 – 20 I 2 = 40 Now we have 2 eq. , 2 unknowns. Add the equations together: -40 I 2 = 42 I 2 = -1. 05 A note that this means direction of I 2 is opposite to that shown on the previous slide Plug into left loop equation: 5 I 1 -10*(-1. 05) = 20 I 1=1. 90 A Use junction equation (eq. 1 from previous page) I 3=I 1+I 2 = 1. 90 -1. 05 I 3 = 0. 85 A