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Question


Given figure shows arrangement of six bulbs and two battries each bulb is rated at 100V and there power is shown in figure
List I gives bulbs and list II gives there output power

With switch S1 closed and switch S2 open

List~IList~III) B1P) 2569II) B2Q) 1289III) B3R) 2009IV) B4S) 5129T) 8009U) 4009

A
I - R, II - U, III - T, IV - R
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B
I - Q, II - S, III - U, IV - T
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C
I - U, II - T, III - U, IV - R
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D
I - R, II - U, III - T, IV - P
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Solution

The correct option is C I - U, II - T, III - U, IV - R
IU,IIT,IIIU,IVR
Since P=V2RR=V2P
Resistance of Bulbs B1,B2,B3 & B4 are R1=1002100=100Ω, R2=100250=200Ω , R100w=100Ω
R3=100225=400, R4=100250=200Ω, R200w=50Ω
With S1 closed S2 open ε=i Req300=i×300
i=1A Power output of bulbs is P=i2R
P1=(23)2×100=4009wP2=(23)2×200=8009w
P3=(13)2×400=4009wP4=(13)2×200=2009w

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