An electric bulb marked as 50 W - 200 V is connected across a 100 V supply- The present power of the bulb is37-5 W10 W12-5 W25 W

Question

Toppr Admin · Accepted Answer

We know that-xA0-R-V2PGiven- V-200-xA0-volt-P-40-xA0-W-xA0-and V-x2032-100-xA0-voltHence- R-V2P-200-xD7-20050Present power- P-x2032-V-x2032-2R-100-xD7-100-xD7-50200-xD7-200P-x2032-12-5-xA0-W

An electric bulb marked as $50 W - 200 V$ is connected across a $100 V$ supply. The present power of the bulb is
$37.5 W$
$10 W$
$12.5 W$
$25 W$

We know that: $R = \frac{V^{2}}{P}$
Given: $V = 200 v o l t, P = 40 W$ and $V^{'} = 100 v o l t$
Hence, $R = \frac{V^{2}}{P} = \frac{200 \times 200}{50}$
Present power: $P^{'} = \frac{V^{' 2}}{R} = \frac{100 \times 100 \times 50}{200 \times 200}$
$P^{'} = 12.5 W$

An electric bulb marked as 50 W−200 V is connected across a 100 V supply. The present power of the bulb is37.5 W10 W12.5 W25 W

We know that: R=V2PGiven: V=200 volt,P=40 W and V′=100 voltHence, R=V2P=200×20050Present power: P′=V′2R=100×100×50200×200P′=12.5 W

An electric bulb marked as $50 W - 200 V$ is connected across a $100 V$ supply. The present power of the bulb is
$37.5 W$
$10 W$
$12.5 W$
$25 W$

We know that: $R = \frac{V^{2}}{P}$
Given: $V = 200 v o l t, P = 40 W$ and $V^{'} = 100 v o l t$
Hence, $R = \frac{V^{2}}{P} = \frac{200 \times 200}{50}$
Present power: $P^{'} = \frac{V^{' 2}}{R} = \frac{100 \times 100 \times 50}{200 \times 200}$
$P^{'} = 12.5 W$