The wiring of a house has resistance 6Ω. A 100W,200V bulb is glowing in the bathroom. A geyser of 1000W,220V is switched on. The drop in potential across the bulb is:
A
0
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B
24V
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C
32V
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D
12V
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Solution
The correct option is C24V Rbulb=2202100=484ΩRGeyser=48.4Ω (i) When only bulb is ON Vbulb=220×484490=217.4V
(ii) When geyser is also switched ON Rnet=484×48.4484+48.4=44Ω; Vbulb=220×4450=193.6V △V=(217.4−193.6)V=24V