The wiring of a house has resistance $6\Omega .$ A 100 W bulb is glowing. If a geyser of 1000 W is switched on, the change in potential drop across the bulb is nearly (Supply voltage is 220 V)

Solution:

${\mathrm{R}}_{\mathrm{Bulb}}=\frac{{220}^2}{100}=48.4\mathrm{\Omega }$

(i) When only bulb is ON,  ${\mathrm{V}}_{\mathrm{Bulb}}=\frac{220\times 484}{490}=217.4\mathrm{V}$

(ii) When geyser is also switched ON, equivalent resistance of bulb and geyser is $\mathrm{R}=\frac{484\times 48.4}{484+48.4}=44\mathrm{\Omega }$

Voltage across the bulb, ${\mathrm{V}}_{\mathrm{Bulb}}=\frac{220\times 44}{50}=193.6\mathrm{V}$

Hence, the potential drop is $217.4-193.6=23.8V$