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.

Answer 1

Energy Dissipated is $2.61 \times 10^-^4$ J

Explanation:

• Energy Dissipated is = $\int_{0}^{t}\frac {E^2}{R}$

t = 1 ms = $10^-^3$s.t

= $\int_{0}^{t}\frac {12 sin(250 \pi t)^2dt}{100}$

• But $(sin^2\theta = \frac {1 - cos^2\theta}{2})$

= $\frac {144}{100}\int_{0}^{t}\frac {1- cos2 \times 250 \pi t}{2}dt$

= $\frac {144}{100}(1 - cos500 \pi t)dt$

• on Integrating we get-

energy dissipated = $\frac {144}{200} [t - \frac {sin500 \pi t}{500 \pi}]$

At t = $10^-^3\ s$

Energy dissipated = $\frac {144}{200} [10^-^3 - \frac {1}{500 \pi}]$

= $\frac {144}{200} [\frac {1}{1000} - \frac {1}{500 \times 3.14}]$

= $2.61 \times 10^-^4$ J