Three electric bulbs rated 40, 60 and 100 watts respectively are connected in parallel with 230 volts d.c. mains. The current (in amp) through the 60 W bulb and the effective resistance (in ohm) of the circuit will be respectively, approximately:

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0.23 A

329 Ω

0.26 A

265 Ω

answer is C.

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Detailed Solution

R=v2P1Req=1V2P1+P2+P3Req=230×23040+60+100=230×230200=265Ω Current in 60W bulb =PV=60230=0.26amp

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