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 changee in potential drop across the bulb is nearly

Options

A. Nil
B. `23 V`
C. `32 V`
D. `12 V`

Answer

B

Explanation

(b) `R_(B u l b) = (220^(2))/(100) = 484 Omega, R_(Geyser) = (220^(2))/(1000) = 48.4 Omega`
(i) When only bulb is `ON`, `V_(B u l b) = (220 xx 484)/(490) = 217.4 V`
(ii) When geyser is also switched `ON`, equivalent resistance of bulb and geyser is `R = (484 xx 48.4)/(484 + 48.4) = 44 Omega`
Voltage across the bulb `V_(B u l b) = (220 xx 44)/(50) = 193.6 V`
Hence, the potential drop is `217.4 - 193.6 = 23.8 V`