Kirchhoffs Laws Kirchhoffs Rules Kirchhoffs Junction Rule Current

Kirchhoff’s Laws

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

At the junction shown: I 1 + I 2 = I 3

Going around the right loop: V 1 – V 2 = I 1 R 1 - I 2 R 1

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 drops Be careful about signs Drop in the direction of current means positive, opposite the direction of current means negative 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: Label currents Choose loop Write KLR e 1 - e 2 = IR 1 + IR 2 50 - 10 = 5 I +15 I I = +2 amps I B 1= 50 V A 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 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.

Calculate the currents. Use outer loop e 1 = I 1 R 1 10 = 10 I 1 R=10 W I 1 E 2 = 5 V I 2 IB E 1 = 10 V I 1= 1 amps Use upper loop e 2 = I 1 R 1 - I 2 R 2 5 = 10(1) - 10 I 2 = 0. 5 amps R=10 W Use lower loop (check) e 1 - e 2 = I 2 R 2 10 - 5 = 10 I 2 = 0. 5 amps

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 does outer loop change when switch is open or closed?

Kirchhoff’s Junction Rule Current Entering = Current Leaving I 1 = I 2 + I 3 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 = I 1 + I 2 = 1. 5 A IB + E 1 = 10 V

Kirchhoff’s Laws (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. 1. 2. 3. Label all currents (Choose any direction) Write down junction equation 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 Loop 2: e 2 = - I 2 R 2 - I 3 R 3 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 Ω I 1 I 3 1. junction: I 3=I 1+I 2 2. left loop: 20 = 5 I 1 -10 I 2 3. right loop: 2 = - 10 I 2 - 10 I 3 I 2 + - 10 Ω 20 V - + 2 V 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 equations and 2 unknowns.

-10 I 1 -20 I 2 = 2 I 1 - 10 I 2 = 20 Substitute and solve: I 2 = -1. 05 A note that this means direction of I 2 is opposite to that shown on the previous slide (Naturally, the current is not negative, but you must keep it negative for the rest of the calcultations) 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