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

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Zero

2.9 volt

10.4 volt

answer is D.

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Detailed Solution

Power of bulb= 60W(given) Resistance of bulb= $= \frac{120 \times 120}{60} = 240 Ω [∵ P = \frac{V^{2}}{R}]$

Power of heater=240W(given) Resistance of heater= $= \frac{120 \times 120}{240} = 60 Ω$

Voltage across bulb before heater is switched on, $V_{1} = \frac{240}{246} \times 120 = 117.073 volt$

Voltage across bulb after heater is switched on, $V_{2} = \frac{48}{54} \times 120 = 106.66 volt$

Hence decrease in voltage $V_{1} - V_{2} = 117.073 - 106.66$ $= 10.4 Volt(approximately)$

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