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Chapter: 12th Physics : Current Electricity

Kirchhoff’s Rules: Solved Example Problems

Solved Example Problems: Kirchhoff’s first rule (Current rule or Junction rule), Kirchhoff’s Second rule (Voltage rule or Loop rule), Wheatstone’s bridge, Meter bridge


Kirchhoff’s first rule (Current rule or Junction rule): Solved Example Problems


EXAMPLE 2.20

From the given circuit find the value of I.


Solution

Applying Kirchoff’s rule to the point P in the circuit,

The arrows pointing towards P are positive and away from P are negative.

Therefore, 0.2A – 0.4A + 0.6A – 0.5A + 0.7A – I = 0

1.5A – 0.9A – I = 0

0.6A – I = 0

I = 0.6 A


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 = ξ 

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  – 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 – I2 ) – 1I2 = 6

3I1 – 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


Meter bridge : Solved Example Problems


EXAMPLE 2.25

In a meter bridge with a standard resistance of 15 Ω in the right gap, the ratio of balancing length is 3:2. Find the value of the other resistance.

Solution


 

EXAMPLE 2.25

In a meter bridge, the value of resistance in the resistance box is 10 Ω. The balancing length is l1 = 55 cm. Find the value of unknown resistance.

Solution

Q = 10 


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12th Physics : Current Electricity : Kirchhoff’s Rules: Solved Example Problems |


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