Kirchhoff’s Rules
Important points to consider:
! Assume and draw a direction for each current.
! Assign and draw signs for each resistor’s ∆V. The current
enters the resistor as (+) and exits as (-).
! Do not change the direction of currents after starting the
problem.
! Take note of the currents entering and leaving any joint: Σ I =
0 at any joint.
! Σ V = 0 around any closed loop.
Kirchhoff’s Rules
! A negative answer simply means that you assumed the wrong
direction, initially, for that current. Do not make any changes in
the middle of the problem. If you substitute that value into
another equation, you must include the negative sign.
The Problem:
Consider the circuit below and solve for each current.
4.0 Ω
4.0 Ω
+
3.0 V
-
4.0 Ω
6.0 Ω
6.0 Ω
4.0 Ω
Let’s mark some points for reference.
4.0 Ω
4.0 Ω
+
3.0 V
-
6.0 Ω
6.0 Ω
4.0 Ω
4.0 Ω
F
C
B
A
E
D
Now, assume current directions and remember that Σ I = 0 at
any joint.
C
B
A
4.0 Ω
4.0 Ω
+
3.0 V
-
6.0 Ω
6.0 Ω
4.0 Ω
4.0 Ω
F
E
D
Draw signs for each resistor’s ∆V. The current enters the
resistor as (+) and exits as (-).
A
_
+
B
+
4.0 Ω
3.0 V
+
6.0 Ω
6.0 Ω
-
C
4.0 Ω
+
+
_
_
_
4.0 Ω
4.0 Ω
F
_
+
_
E
+
D
The currents need a name.
I1
A
I2
_
+
B
+
4.0 Ω
3.0 V
6.0 Ω
-
C
4.0 Ω
+
+
_
+
6.0 Ω
I3
_
_
4.0 Ω
4.0 Ω
F
_
+
I1
_
+
E
I2
D
Sum the potentials (voltages) around any loop.
Loop A-F clockwise beginning at ‘A’:
− 4 I1 − 4 I 2 − 6 I 2 − 4 I 2 − 4 I1 + 3 V = 0
Equation # 1:
+ 8 I1 + 14 I 2 = + 3 V
Loop A-B-E-F clockwise beginning at ‘A’:
− 4 I1 − 6 I 3 − 4 I1 + 3 V = 0
+ 4 I1 + 6 I 3 + 4 I1 = + 3 V
Equation # 2:
+ 8 I1 + 6 I 3 = + 3 V
Notice this relationship based upon
this particular problem:
I 1=I
2
+I
3
Rename I3 in Equation # 2:
+ 8 I1 + 6 I 3 = + 3 V
+ 8 I1 + 6 ( I 1 − I
2
+ 8 I1 + 6 I 1 − 6 I
Equation # 3:
14 I1 − 6 I
2
) = + 3V
2
= + 3V
= + 3V
Now, solve Equation # 3 for I 1 and substitute I 1 into Equation # 1.
14 I1 − 6 I
2
= + 3V
3V + 6 I
I1 =
14
2
It is interesting to note, here, that we are summing voltages,
so 14 I 1 has units of volts, and 14 has units of resistance ( Ω ).
3V + 6 I
I1 =
14
2
Equation # 1:
+ 8 I1 + 14 I 2 = + 3 V
3V + 6 I 2 
I
V
+8
+
=
+
14
3
2

14


24 V + 48 I
14
24 V 48 I
+
14
14
2
2
+ 14 I 2 = 3 V
+ 14 I 2 = 3 V
1.7 V + 3.4 I
2
+ 14 I 2 = 3 V
17 I 2 = 1.3 V
−3
I 2 = 76 [10 ] A = 76 m A = I 2
Equation # 1:
+ 8 I1 + 14 I 2 = + 3 V
+ 8 I1 = + 3 V − 1.1 V = 1.9 V
1.9 V
I1 =
= 0.24 A = I1
8
And, from earlier:
I 1=I
I
I
3
3
+I
3
=I 1−I
2
2
= 0.24 A − 76 m A = 0.16 A = I
3