Text section 28.3
Kirchhoff’s circuit rules
Practice:
Chapter 28, problems 17, 19, 25, 26, 43
Junction Rule: total current in = total current out
at each junction (from conservation
of charge).
Loop Rule: Sum of emfs and potential differences
around any closed loop is zero (from
conservation of energy).
1
Junction Rule: conservation of charge.
I2
I1
I 1 = I2 + I 3
I3
or
I 1’
I2’ (= -I2)
I 1’ + I 2 ’ + I 3 ’ = 0
I3’ (= -I3)
Loop Rule: conservation of energy.
Follow a test charge q around a loop:
around any loop in
circuit.

R
I
R
I
-
+
-Q
+Q
changes going from left to right
ΔV = -IR
ΔV = +IR
ΔV =
ΔV = Q/C
C
2
18 Ω
9V
b
6Ω
I2
I1
3V
a
Find the current
through each battery.
c
3Ω
I3
d
18 Ω
The junction rule
will give:
A)  I1 + I2 + I3 = 0
B)  -I1 + I2 + I3 = 0
C)  I1 - I2 + I3 = 0
D)  I1 + I2 - I3 = 0
E) none of these
9V
b
6Ω
I2
I1
3V
a
c
3Ω
I3
d
3
18 Ω
9V
b
6Ω
I2
I1
3V
a
Find the current
through each battery.
c
3Ω
I3 = - (I1 + I2)
d
18 Ω
The junction rule
will give:
A)  I1 + I2 + I3 = 0
B)  -I1 + I2 + I3 = 0
C)  I1 - I2 + I3 = 0
D)  I1 + I2 - I3 = 0
E) none of these
9V
b
6Ω
I2
I1
3V
a
c
3Ω
I3
d
4
18 Ω
The loop rule applied to
loop abcda will give:
9V
b
6Ω
I2
I1
c
3Ω
3V
a
I3
d
A)  9A - 18I1 -3I3 = 0
B)  9A + 18I1 -3I3 = 0
C)  9A + 18I1 +3I3 = 0
D)  9A - 18I1 +3I3 = 0
E) none of these
18 Ω
The loop rule applied to
loop abda will give:
9V
b
6Ω
I2
I1
3V
a
c
3Ω
I3
d
A)  12A - 18I1 + 6I2 = 0
B)  12A - 18I1 - 6I2 = 0
C)  6A - 18I1 - 6I2 = 0
D)  6A + 18I1 + 6I2 = 0
E)  6A - 18I1 + 6I2 = 0
5
18 Ω
9V
b
6Ω
I2
I1
3Ω
3V
a
Find the current
through each battery.
c
I3 = - (I1 + I2)
d
Answers: I1 = 400 mA, I2 = 200 mA
18 Ω
9V
b
I2
I1
c
6Ω
3V
a
3Ω
I3 = - (I1 + I2)
d
Solution:
Loop abcda:
..... 1
Loop dbcd:
7× 2 - 1 : 60 I2 = 12A
Junction b:
I 1 + I2 + I3 = 0
so I3 = -(I1 + I2)
..... 2
 I2 = 200 mA
3 × 1 - 2 : 60 I1 = 24A 
I1 = 400 mA
6
10 Ω
6V
b
20 Ω
I2
I1
I3 = - (I1 + I2)
b
c
20 Ω
I2
a
Find the current
through each battery.
d
10 Ω
I1
40 Ω
3V
a
6V
c
3V
40 Ω
Find the current
through each battery.
I3 = - (I1 + I2)
d
Answers: I1 = 171 mA, I2 = -64.3 mA
7
10 Ω
6V
b
I2
I1
c
20 Ω
40 Ω
3V
a
Solution:
Loop abcda:
I3 = - (I1 + I2)
Junction b:
I 1 + I2 + I3 = 0
so I3 = -(I1 + I2)
d
..... 1
Loop dbcd:
..... 2
4x 1 - 5x 2 : -140 I2 = 9
 I2 = -64 mA
3x 1 - 2x 2 : 70 I1 = 12  I1 = 171 mA
c
12 V
200 Ω
200Ω
a
b
600 Ω
300Ω
d
What is Vab (i.e., Va - Vb ) when the switch is open?
Exercise for fun: Find the current through the
switch when it is closed.
8
The loop rule requires
that Vab (i.e., Va - Vb )
200 Ω
I1
c
200Ω
a
obeys:
b
600 Ω
A)
B)
C)
D)
Vab = (200Ω)I1 + (200Ω)I2
Vab = (200Ω)I1 - (200Ω)I2
Vab = -(200Ω)I1 + (200Ω)I2
Vab = -(200Ω)I1 - (200Ω)I2
I2
300Ω
d
c
2Ω
1Ω
Reff = ?
a
1Ω
2Ω
b
1Ω
d
“Series and parallel“ rules don’t help in this case. You
have to go back to the fundamentals—Kirchhoff’s
Circuit Rules.
(Answer: Reff = 1.4Ω )
9
Reff = VTOTAL/ITOTAL
ITOTAL
c
I1
VTOTAL
1Ω
a
I4
I3
2Ω
b
1Ω
2Ω
d
Show that:
I2
I5
1Ω
I1 = I 5 = 3I 3
I 2 = I 4 = 2I 3
ITOTAL= 5I3
VTOTAL= (7Ω) I3
1)  Use Kirchhoff’s rules to write everything in terms
of, one variable, e.g., I3.
2)  Divide VTOTAL/ITOTAL .
Reff = VTOTAL/ITOTAL
ITOTAL
c
I1
VTOTAL
1Ω
a
I4
I3
2Ω
b
1Ω
2Ω
d
Show that:
I2
1Ω
I5
I1 = I 5 = 3I 3
I 2 = I 4 = 2I 3
ITOTAL= 5I3
VTOTAL= (7Ω) I3
1)  Use Kirchhoff’s rules to write everything in terms
of, e.g., I3. (Results: I1 = I5 = 3I3, I2 = I4 = 2I3; ITOTAL= 5I3;
VTOTAL= (7Ω) I3.)
2)  Divide VTOTAL/ITOTAL .
10
Solution to part A:
c
250
12 V
I1
a
I1 + I 3
Solution to part b:
I2
100
I3
b
Loop adba:
400
250
d
I1 - I3
1
Loop abca:
2
Then:
 Solve…
11
R1
V1
V
R2
V2
and
(from
; and
etc)
c
3Ω
1Ω
Reff = ?
a
5Ω
2Ω
b
4Ω
d
Notice that “series and parallel“ don’t help in this
case. You have to go back to Kirchhoff’s Rules.
(Answer: Reff = ( 155/74 ) Ω )
12