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Question

A resistor of resistance 100 Ω is connected to an AC source ε = (12 V) sin (250 π s−1)t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.

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Solution

Given:
Peak voltage of AC source, E0 = 12 V
Angular frequency, ω = 250 π s−1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat H is given by,
H=Erms2RT
Here, Erms = RMS value of voltage
R = Resistance of the resistor
T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,
H=0tdH =E02 sin2 ωtRdt Erms =E0sinωt = 144100010-3sin2 ωt dt = 1.44 010-31-cos 2 ωt2dt =1.442010-3dt+010-3cos 2 ωt dt =0.72 10-3-sin2ωt2ω010-3 = 0.7211000-1500 π = 0.7211000-21000π = π-21000 π×0.72 = 0.0002614 = 2.61×10-4 J

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