First slide

Electrical power

Question

The wiring of a house has resistance $6 Ω .$ 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)

Moderate

A

nil
B

23 V
C

32 V
D

12 V

Solution

$R_{Bulb} = \frac{220^{2}}{100} = 48 .4 Ω$
(i) When only bulb is ON, $V_{Bulb} = \frac{220 \times 484}{490} = 217 .4 V$
(ii) When geyser is also switched ON, equivalent resistance of bulb and geyser is $R = \frac{484 \times 48 .4}{484 + 48 .4} = 44 Ω$
Voltage across the bulb, $V_{Bulb} = \frac{220 \times 44}{50} = 193 .6 V$
Hence, the potential drop is $217.4 - 193.6 = 23.8 V$

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