Solved Example Problems [Kirchoff's Laws]

KIRCHOFF'S LAWS

Example: 16

Find the current through 12 Ω resistor as shown in figure.

Incoming current at point A is

5 + 6 = 11 A

The current through 12 Ω resistor is 11 A.

Example: 17

Find the voltage across the resistor 'R' in figure.

Applying KVL in the above circuit

20 - 8 + 10 - V1 – 17 + 5 = 0

10 – V1 = 0

10 = V1

Voltage across the resistor R = 10 V

Example: 18

For the circuit shown in figure, determine the unknown voltage drop V1,

Applying KVL in the above circuit

-2  -1 - V1 - 3 - 5 + 30 = 0

-V1 + 19 = 0

19 = V1

V1 = 19 V

Unknown voltage drop V1

Example: 19

Determine the current through resistance R3 in the circuit shown in figure.

According to Kirchoff's current law

IT = I1 + I2 + 13

50 = 30 + 10 + I3

I3 = 50 - 40

I3 = 10 mA

Example: 20

Find the current I and the voltage across 30 Ω shown in the figure.

Voltage drop across each resistor as

V8 = 8I, V30 = 30I, V2 = 2I

Applying KVL

100 - 40 = 8I + 30I + 2I

60 = 40I

I = 60/40 = 1.5 A

Voltage across 30 Ω = V30 = 30 I = 30 x 1.5 = 45 V

Example: 21

For the resistive circuit shown in fig. Find (a) I1 (b) VS.

(a) Applying KCL at node 1,

I1 = 4 – 1 = 3A

(b) V1 = 3 × 1 = 3V

V12 = 1 × 1 = 1V

V2 = 3 -1= 2V

I2 = V2/1 = 2/1 = 2A

KCL at node 2

I 3 = 1- I2 = 1- 2 = -1 A

VS = V2 - ( 1 × I3 ) = 2 - (1×-1)

=2+1=3V

Example: 22

Using KCL, find V2 for the given figure and given values i1 = 4A , V3 = 3V, V4 = 8V.

Applying KCL at the node

i1 – 3 i2 + i2 – i3 – i4 = 0

i1 – 2 i2 – i3 – i4 = 0

Substituting the values

4 - 2i2 - 1 – 2 = 0

1 – 2 i2 = 0

1 = 2 i2

i2 = 1/2 = 0.5 A

V2 = i2 × R = 0.5 x 2 = 1V