Physics 1161: Lecture 10 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.

Using Kirchhoff’s Rules (1)Label all currents (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 summing from negative to positive For resistors – voltage change is negative when summing in the direction of the current R4R4 I1I1 I3I3 I2I2 I4I4 R1R1 11 R2R2 R3R3  2 2 33 R5R5 A B (2) Write down junction equation I in = I out I5I5

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

Loop Rule 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 Find I: Label currents Choose loop Write KLR

Resistors R 1 and R 2 are 1.In parallel 2.In series 3.neither I1I1 R 1 =10  R 2 =10  E 1 = 10 V IBIB E 2 = 5 V I2I2 + -

Resistors R 1 and R 2 are 1.In parallel 2.In series 3.neither I1I1 R 1 =10  R 2 =10  E 1 = 10 V IBIB E 2 = 5 V I2I2 + - Definition of parallel: 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 R=10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R=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 = 0 24% 62% 24% Calculate the current through resistor  I 1 = E 1 /R = 1A

How would I 1 change if the switch was opened? E 1 = 10 V IBIB R=10  I1I1 I2I2 E 2 = 5 V 1.Increase 2.No change 3.Decrease

How would I 1 change if the switch was opened? E 1 = 10 V IBIB R=10  I1I1 I2I2 E 2 = 5 V 1.Increase 2.No change 3.Decrease

Preflight 10.2 R=10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R=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 = 0 43%  I 2 = 0.5A Calculate the current through resistor %

Preflight 10.2 R=10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R=10  I2I E 1 - E 2 + I 2 R = 0 Note the sign change from last slide  I 2 = -0.5A Answer has same magnitude as before but opposite sign. That means current goes to the left, as we found before. 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. Work through preflight with opposite sign for I 2 ?

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 7% 37% 57% R=10  E 1 = 10 V IBIB I1I1 E = 5 V R=10  I2I2 + - Preflight 8.3 “The first two can be calculated using V=IR because the voltage and resistance is given, and the current through E1 can be calculated with the help of Kirchhoff's Junction rule, that states whatever current flows into the junction must flow out. So I1 and I2 are added together.”

Kirchhoff’s Laws (1)Label all currents Choose any direction (2)Write down the junction equation I in = I out (3)Choose loop and direction Your choice! (4)Write down voltage changes Follow any loops (5)Solve the equations by substitution or combination. R4R4 R1R1 E1E1 R2R2 R3R3 E2E2 E3E3 I1I1 I3I3 I2I2 I4I4 R5R5 A B

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. R1R1 R2R2 R3R3 I1I1 I3I3 I2I2 + - 11 22 + -

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) 3. Choose loop and direction (Your choice!) 4.Write down voltage changes Loop 2: 11 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   2. Write down junction equation

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 10 I1I1 I3I3 I2I2 + -  junction: I 3 =I 1 +I 2 2. left loop: I 1 +10I 2 = 0 3. right loop: I I 3 = 0 solution: substitute Eq.1 for I 3 in Eq. 3: rearrange: -10I I 2 = 2 rearrange Eq. 2: 5I 1 -10I 2 = 20 Now we have 2 eq., 2 unknowns. Continue on next slide

-10I 1 -20I 2 = 2 2*(5I I 2 = 20) = 10I 1 – 20I 2 = 40 Now we have 2 eq., 2 unknowns. Add the equations together: -40I 2 = 42 I 2 = A note that this means direction of I 2 is opposite to that shown on the previous slide Plug into left loop equation: 5I 1 -10*(-1.05) = 20 I 1 =1.90 A Use junction equation (eq. 1 from previous page) I 3 =I 1 +I 2 = I 3 = 0.85 A