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