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Q.

Three 60 W, 120 V light bulbs are connected across 12o V power line as shown in fig. The total power dissipated in the three bulbs is

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a

60W

b

180w

c

90 W'

d

40 W

answer is D.

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Detailed Solution

The resistance of each bulb is given byR=V2P=(120)260=240ΩTotal resistance of the circuitR=R+R×RR+R=240+240×240240+240=360ΩThe current in the circuiti=VR=120360=13ampTotal power produced isP=i2R=132×360=40W
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