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प्रश्न
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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उत्तर
Given:
Peak voltage of AC source, E0 = 12 V
Angular frequency, ω = 250 πs−1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat (H) is given by,
`H = E_{rms}/R T`
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 = ∫_0^t dH`
= `∫ (E_0^2 sin^2 omega t)/R dt (therefore E_{rms = E_0 sin omega t})`
= `144/100 ∫_0^{10-3} sin^2 omegat dt`
=` 1.44/2 ∫_0^{10-3} ((1-cos 2 omegat)/2)dt`
=`1.44/2 [ ∫_0^{10-3} dt + ∫_0^{10-3} cos 2 omega t dt]`
= `0.72 [ 10^-3 - {(sin 2 omegat)/(2 omega) }]_0^(10-3)]`
`= 0.72[1/1000 - 1/(1000pi)]`
`=0.72 [1/1000 - 2/(1000pi)]`
`= ((pi - 2)/(1000pi)) xx 0.72`
`= 0.0002614 = 2.61 xx 10^-4 J`
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