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Question

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

A
zero volt
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B
2.9 volt
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C
13.3 volt
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D
10.4 volt
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Solution

The correct option is D 10.4 volt
Resistance of bulb =120×12060=240Ω
Resistance of heater =120×120240=60Ω
Voltage across bulb before heater switched, on
V1=12024×240=117.07
Voltage across bulb after heater switched on V2
V2=12054×48=100.60
Decrease of water =V1V2=10.04

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