Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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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. - Physics

Sum

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 `pi` s−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 omega)/(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`

Concept: Peak and Rms Value of Alternating Current Or Voltage
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 17 Alternating Current
Q 11 | Page 330
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