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The resistances of the four arms P, Q, R and S in 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. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be:
A
2.0 A
B
1.0 A
C
0.2 A
D
0.1 A
Detailed Solution
$\frac{P}{Q} = \frac{R}{S} = \frac{1}{3}$, so the bridge is balanced and no current flows through the galvanometer.
Resistance of the bridge $= \frac{(40)(120)}{40 + 120} = 30\ \Omega$
Current through the cell $= \frac{7}{5 + 30} = \frac{1}{5}$ A = 0.2 A
Resistance of the bridge $= \frac{(40)(120)}{40 + 120} = 30\ \Omega$
Current through the cell $= \frac{7}{5 + 30} = \frac{1}{5}$ A = 0.2 A