The supply voltage to room is 120 V. The resistance of the lead wires… | JEE Main 2013

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Electrical Power & Heating Effect

The supply voltage to room is $120\text{ V}.$ The resistance of the lead wires is $6\ \Omega$ . A $60\text{ W}$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $240\text{ W}$ heater is switched on in parallel to the bulb?

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Answer: (d)

$R_{bulb} = \frac{120^2}{60} = 240\ \Omega$ , $R_{heater} = \frac{120^2}{240} = 60\ \Omega$ .
Before: $V_1 = 120 \times \frac{240}{240 + 6} \approx 117.1$ V.
After: bulb $\parallel$ heater $= \frac{240 \times 60}{300} = 48\ \Omega$ , so $V_2 = 120 \times \frac{48}{48 + 6} \approx 106.7$ V.
Decrease $= 117.1 - 106.7 \approx 10.4$ V. So the answer is (d).

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