Two bulbs are marked 100 W, 220 V and 60 W, 110 V. Calculate
the ratio of their resistances.
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Power, P=V2R
R1= 220×220/50
=968Ω
R2 = 220×220/100
=484Ω
The equivalent resistance in the circuit,
Req =R1 +R2
Req =968+484=1452Ω
The current flow in the circuit,
I= 440/1452
=0.3030A
The voltage drop across bulb 1.
V1=iR1 =0.3030×968=293.33V
The voltage drop across bulb 2.
V2=iR2=0.3030×484=146.65V
Bulb 1 will be fused because it can tolerate only 220 V but the voltage drop across it is 293.33V
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