The resistances of the four arms P, Q, Rand 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 7volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be

The situation is as shown in the figure.

For a balanced Wheatstone's bridge

$$\begin{gathered} \frac{P}{Q}=\frac{R}{S} \\ \therefore \frac{10\Omega }{30\Omega }=\frac{30\Omega }{90\Omega }or\frac{1}{3}=\frac{1}{3} \end{gathered}$$

It is a balanced Wheatstone's bridge. Hence no current flows in the galvanometer arm. Hence, resistance 50 Q becomes ineffective.

$\therefore$The equivalent resistance of the circuit is

$$\begin{gathered} R_{eq}=5\Omega +\frac{\left(10\Omega +30\Omega \right)\left(30\Omega +90\Omega \right)}{\left(10\Omega +30\Omega \right)+\left(30\Omega +90\Omega \right)} \\ =5\Omega +\frac{\left(40\Omega \right)\left(120\Omega \right)}{40\Omega +120\Omega }=5\Omega +30\Omega =35\Omega \end{gathered}$$

Current drawn from the cell is

$I=\frac{7V}{35\Omega }=\frac{1}{5}A=0.2A$