Question

The wiring of a house has resistance $6\Omega$. A $100W,200V$ bulb is glowing in the bathroom. A geyser of $1000W,220V$ is switched on. The drop in potential across the bulb is:

• A. $0$
• B. $24V$
• C. $32V$
• D. $12V$

Answer 1

Answer: (B)

$24V$

$R_{bulb}=\frac{{220}^2}{100}=484\Omega$ $R_{Geyser}=48.4\Omega$
(i) When only bulb is ON
$V_{bulb}=\frac{220\times 484}{490}=217.4V$

(ii) When geyser is also switched ON
$R_{net}=\frac{484\times 48.4}{484+48.4}=44\Omega$;
$V_{bulb}=\frac{220\times 44}{50}=193.6V$
$△V=(217.4-193.6)V=24V$