Solved Example Problems: Kirchhoffâ€™s first rule (Current rule or Junction rule), Kirchhoffâ€™s Second rule (Voltage rule or Loop rule), Wheatstoneâ€™s bridge, Meter bridge

EXAMPLE 2.20

From the given circuit find the value of I.

Solution

Applying Kirchoffâ€™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 â€“ 0.4A + 0.6A â€“ 0.5A + 0.7A â€“ I = 0

1.5A â€“ 0.9A â€“ I = 0

0.6A â€“ I = 0

I = 0.6 A

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 Kirchoffâ€™s voltage rule.

*Solution*

Thus applying Kirchoffâ€™s second law to the closed loop EACE

*I*1*R*1* *+* I*2*R*2* *+* I*3*R*3* *=* *Î¾

and for the closed loop ABCA

*I*4*R*4* *+* I*5*R*5-*I*2*R*2= 0

EXAMPLE 2.22

Calculate the current that flows in the 1â€†Î© 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 Kirchoffâ€™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 ) â€“ 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.

EXAMPLE 2.23

In a Wheatstoneâ€™s bridge P = 100 Î©, Q = 1000 Î© and R = 40 Î©. If the galvanometer shows zero deflection, determine the value of S.

*Solution*

EXAMPLE 2.24

What is the value of *x* when the Wheatstoneâ€™s network is balanced?

P = 500 Î©, Q = 800 Î©, R = *x* + 400, S = 1000 Î©

*Solution*

P/Q = R/S

x + 400 = 0.625 Ã— 1000

x + 400 = 625

x = 625 â€“ 400

x = 225 Î©

EXAMPLE 2.25

In a meter bridge with a standard resistance of 15 Î© 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 Î©. The balancing length is *l*1 = 55 cm. Find the value of unknown resistance.

*Solution*

Q = 10 â„¦

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12th Physics : Current Electricity : Kirchhoffâ€™s Rules: Solved Example Problems |

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