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