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

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1.0 A

2.0 A

0.1 A

0.2 A

answer is B.

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Detailed Solution

The given circuit is a balanced wheats tone’s bridge circuit hence galvanometer resistance will be ineffective hence total effective resistance of the circuit
$\begin{array}{l} R_{eff} = r + \frac{(P + Q) (R + S)}{P + Q + R + S} \\ = 5 + \frac{(10 + 30) (30 + 90)}{10 + 30 + 30 + 90} = 35 Ω \\ ∴ i = \frac{V}{R_{eff}} = \frac{7}{35} = 0 .2 A \end{array}$