Kirchhoff’s Rules: Solved Example Problems

Kirchhoff’s Second rule (Voltage rule or Loop rule) : Solved Example Problems

EXAMPLE 2.21

The following figure shows a complex network of conductors which can be divided into two closed loops like ACE and ABC. Apply Kirchoff’s voltage rule.

Solution

Thus applying Kirchoff’s second law to the closed loop EACE 

I1R1 + I2R2 + I3R3 = Î¾ 

and for the closed loop ABCA

I4R4 + I5R5-I2R2= 0

EXAMPLE 2.22

Calculate the current that flows in the 1 Ω resistor in the following circuit.

Solution

We can denote the current that flows from 9V battery as I1 and it splits into I2 and I1  â€“ I2 in the junction according Kirchoff’s current rule (KCR). It is shown below.

Now consider the loop EFCBE and apply KVR, we get

1I2 + 3I1 + 2I1 = 9

5I1 + I2 = 9 (1)

Applying KVR to the loop EADFE, we get

3 (I1 â€“ I2 ) – 1I2 = 6

3I1 â€“ 4I2 = 6 (2)

Solving equation (1) and (2), we get

I1 = 1.83 A and I2 = -0.13 A

It implies that the current in the 1 ohm resistor flows from F to E.

Wheatstone’s bridge : Solved Example Problems

EXAMPLE 2.23

In a Wheatstone’s bridge P = 100 Ω, Q = 1000 Ω and R = 40 Ω. If the galvanometer shows zero deflection, determine the value of S.

Solution

EXAMPLE 2.24

What is the value of x when the Wheatstone’s network is balanced?

P = 500 Ω, Q = 800 Ω, R = x + 400, S = 1000 Ω

Solution

P/Q = R/S

x + 400 = 0.625 × 1000

x + 400 = 625

x = 625 – 400

x = 225 Ω