PHYS 222
Worksheet 12 – Kirchhoff’s Laws
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
2/12/12
Useful Equations
I 0
V  0
Diagrams
Related Problems
Junction Rule. The algebraic sum of the currents into any
junction is zero.
Loop Rule. The algebraic sum of the potential difference
in any loop must equal zero.
1) Find the following for the circuit (Book 26.21)
(Dots on the circuit represent starting points for the loop method)
(a) Current in resistor R
Using the junction rule at junction (a) :
64 I  0
I 2 A
(b) Resistance R
Using loop rule on loop (1) :
2R  28  3(6)  0
R 5 
(c) Unknown EMF
Using loop rule on loop (2) :
6(4)    3(6)  0
   42 V
(d) If the circuit is broken at point x, what is the current in resistor R?
Since the middle branch is broken, we can use the loop rule on loop (1) again, but solving
for a new current:
5I ' 28  3I '  0
 I '  3.5 A
2) Both meters are ideal. Voltmeter reads 16.0 V when switch is open. Find the EMF. What
will the ammeter read when the switch is closed? (Book 26.27)
By junction rule at junction (a) :
16
 0.32 A
50
 I1  I 2  0.32  0
I3 
Using loop rule on loop (1) :
   20I2  30(0.32)  50(0.32)  0
Using loop rule on loop (2) :
 75I1  30(0.32)  50(0.32)  0
Now there are three equations with three unknowns. Solve for each variable:
I1  0.3413 A
I 2  0.6613 A
  28.82 V
When switch is closed, using loop (3) the ammeter reads:
25  50I3 '  0
 I3 '  0.50 A
3) All meters are idealized and the batteries have no appreciable internal resistance. (Book
26.29)
(a) With the switch open, what is the reading of the voltmeter?
Using loop (1) :
25 100I 15  75I  0
 I  0.2286 A
Vba  Vb Va  0.2286(75) 15  2.145 V
(b) With the switch closed, what is the reading of the voltmeter?
When the switch is close, the middle branch becomes a straight wire (short circuit),
meaning the voltage across it is zero. Since the right branch is in parallel with the middle
branch, it too has a voltage of zero:
V=0V
(c) With the switch closed, find the reading of the ammeter. What is the direction of the
current?
Using the junction rule at junction (a) :
 I1  I 2  I '  0
Using loop (2) :
 25 100I1  0
Using loop (1) again with a different current in the right branch:
 25 100I1 15  75I '  0
Now there are three equations with three unknowns. Solve for each variable:
I1  0.25 A
I '  0.2 A
I 2  0.05 A
4) Find the three currents I1, I2, I3 . (Book 26.61)
Using loop (1) :
 12 1I2  5(I2  I3 )  0
Using loop (2) :
 1I1  9  8  I1  I3   0
Using loop (3) :
 10I3  9 1I1 1I2 12  0
Now there are three equations with three unknowns. Solve for each variable:
I1  0.848 A
I 2  2.14 A
I3  0.171 A