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Q.

Three identical bulbs B1,B2 and B3 are connected in series parallel combination as shown. Rating of each bulb is 45 W, 100 Volt. If a 100V supply is connected across the combination, the power consumed by the system will be

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a

160 W

b

90 W

c

180 W

d

30 W

answer is D.

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Detailed Solution

If R be the resistance of each bulb, then R=V2PNow, RC=R×RR+R+R=3R2i=current flowing through B3=V3R/2=23VRP1=i22.R×2+i2R=32i2R=322V3R2R=23V2R=23P∴ P1=23×45W=30W
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