A light bulb is rated 100w for 220v a 220v supply.Find the peak voltage of the sourse.
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Rating of light bulb = 100W-220V
So , it's resistance = } R = V²/P
=} R = 220x220/100
=} R = 484Ω
Now P= V²/R
So , P = 220x220/484
=} P= 100W
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Answer:
The peak voltage of the source is 0.45A
Supply of the bulb = 220V (Given)
Power = 100 watt (Given)
Power of the bulb P = VI
P = V x V/R
R = V² / P
= 220 x 220 / 100
= 48400 / 100
= 484
Peak voltage of the source = E/√2
E0 = 220 × √2
= 220√2
= 311.1
Irms = I0/√2 = E0/R√2
= 311.1/484√2
= 311.1/684.4
= 0.45A
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