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.

#### 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`