First slide

KVL

Question

The resistances of the four arms P, Q, Rand Sin a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. 1f the galvanometer resistance is 50 ohm, the current drawn from the cell will be

Moderate

A

0.1 A
B

2.0 A
C

1.0 A
D

0.2 A

Solution

The situation is as shown in the figure.
For a balanced Wheatstone's bridge
$\frac{P}{Q} = \frac{R}{S} ∴ \frac{10 Ω}{30 Ω} = \frac{30 Ω}{90 Ω} o r \frac{1}{3} = \frac{1}{3}$
It is a balanced Wheatstone's bridge. Hence no current flows in the galvanometer arm. Hence, resistance 50 Q becomes ineffective.
$∴$ The equivalent resistance of the circuit is
$R_{e q} = 5 Ω + \frac{(10 Ω + 30 Ω) (30 Ω + 90 Ω)}{(10 Ω + 30 Ω) + (30 Ω + 90 Ω)} = 5 Ω + \frac{(40 Ω) (120 Ω)}{40 Ω + 120 Ω} = 5 Ω + 30 Ω = 35 Ω$
Current drawn from the cell is
$I = \frac{7 V}{35 Ω} = \frac{1}{5} A = 0.2 A$

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