Kirchhoff’s Laws
Kirchhoff’s laws consist of two statements,
Current Law
At any junction (node) in an electric circuit the total current flowing towards
that junction is equal to the total current flowing away from the junction.
Voltage Law
In any closed loop in a network, the algebraic sum of the voltage drops taken
around the loop is equal to the resultant emf acting in that loop.

Kirchhoff’s Laws
Current Law
At any junction (node) in an electric circuit the total current flowing towards
that junction is equal to the total current flowing away from the junction.
or
The algebraic sum of currents at a node is zero
I1 + I2 = I3 + I4 + I5
or
I1 + I2 – (I3 + I4 + I5 )=0
I3
I4
I5
I2
I1

Kirchhoff’s Current Law
Activity
1. Write the formula and deduce the current I1 for the circuit shown given that
I2 = 2A, I3 = 1A, I4 = 3A and I5 = 4A.
I3
I4
I5
I2
I1
2. Determine the unknown currents.
I1
I4
I5
I3
EC
B
F
D
A
40A
120A
50A
20A 15A
I2

Kirchhoff’s Laws
Voltage Law
In any complete electric circuit, the algebraic sum of the source voltage must
equal the algebraic sum of the voltage drops.
or
In any closed loop, the algebraic sum of all voltages is zero
-
I
+
E
+ + +- - -
R2R1 R3
L1
VR1 + VR2 + VR3 = EE - VR1 - VR2 - VR3 = 0 or
VR1 = IR1 VR2 = IR2 and VR3 = IR3
IR1 + IR2 + IR3 = E

Kirchhoff’s Laws
Voltage Law
Example of a combinational circuit.
L2
+
-
E
+
-
R2
+ -
R1
+
-
R3
I1 + I2
L1
I2
I1
VR1 + VR2 = E
E - VR1 - VR2 = 0
IR1 + IR2 = E . . . . . .(1)
Loop 1
VR1 + VR3 = E
E - VR1 – VR3 = 0
Loop 2
IR1 + IR3 = E . . . . . .(2)
Equations (1) and (2) are then solved simultaneously

Kirchhoff’s Voltage Law
Activity
Generate the voltage equations and show that all voltages sum to zero for the
circuits shown. Determine all network currents.
Label direction of current flow and polarity at terminals of all resistors
Trace your loops in a clockwise direction
(direction of conventional current flow)
-
+
20V
R1 = 2Ω
R3 = 4Ω
R2 = 6Ω
-
+
10V
R1 = 4Ω
R3 = 2ΩR2 = 6Ω
circuit 1 circuit 2

Kirchhoff’s Laws
Voltage Law
Example of a dual source network.
Equations (1) and (2) are then solved simultaneously
+
-
E1
+
-
R3
+ -
R1
I1 + I2
L1
I2
I1
+
-
E2
L2
+-
R2
VR1 + VR3 = E1
E1 - VR1 – VR3 = 0
I1R1 + R3(I1+I2 ) = E1 . . . . . .(1)
Loop 1
VR2 + VR3 = E2
E2 – VR2 – VR3 = 0
Loop 2
I2R2 + R3(I1+I2 ) = E2 . . . . . .(2)

Kirchhoff’s Voltage Law
Activity
1. Determine the network currents and the power dissipated in the
load resistor RL.
RL = 12Ω
R1 = 3Ω R2 = 2Ω
-
+
6VE1 E2
-
+
4V
Label directions of current flow and polarity at terminals of all resistors
Trace your loops following the direction of conventional current
2. After a period of time E2 becomes discharged to 2 volts. Determine the
network currents under these conditions and state reasons.

Kirchhoff’s Voltage Law
Activity
A more practical use of Kirchhoff’s law is that of vehicle electrics. The battery
when fully charged has an open circuit voltage of 12.6 volts and an internal
resistance of 0.1Ω. The alternator provides a dc (rectified) output of 16 volts
and is used to supply current to both the load and battery thus maintaining
a charge on the battery. Determine the current in the load and its source.
Investigate this further to see when the battery begins to take a charge from the
generator using computer simulation.
RLOAD
1.2Ω
Rgen
0.2Ω
-
+
EGEN 16V EBAT
12.6V
Rbat
0.1Ω

Kirchhoff’s Voltage Law
Activity
A more practical use of Kirchhoff’s law is that of vehicle electrics. The battery
when fully charged has an open circuit voltage of 12.6 volts and an internal
resistance of 0.1Ω. The alternator provides a dc (rectified) output of 16 volts
and is used to supply current to both the load and battery thus maintaining
a charge on the battery. Determine the current in the load and its source.
Investigate this further to see when the battery begins to take a charge from the
generator using computer simulation.
RLOAD
1.2Ω
Rgen
0.2Ω
-
+
EGEN 16V EBAT
12.6V
Rbat
0.1Ω