Q80 / 90 · JEE Main 2014 (19 Apr Online) · Physics

Four bulbs $B_{1}$ , $B_{2}$ , $B_{3}$ and $B_{4}$ of $100 W$ each are connected to $220 V$ main as shown in the figure. The reading in an ideal ammeter will be

Solution

The resistance of any bulb $= V^{2} / P$

Current through ammeter at shown position $= 3 i_{1}$

current in any $1$ bulb $= i_{1} = \frac{V}{R} = \frac{V}{V^{2} / P}$ $= \frac{P}{V} = \frac{100}{220}$

Through ammeter, $\Rightarrow i_{net} = 3 i_{1} = \frac{300}{220}$ $= 1 .35 A$

Hence, the correct answer is option: 'B'.