Question

A resistor of resistance $100ohm$ is connected to an AC source, $\left(12V\right)\sin (250\pi s-1)t$. Find the energy dissipated as heat during$t=0$to $t=1$ $ms.$

Answer 1

Given data

1. The AC voltage source is $E=\left(12V\right)\sin (250\pi s-1)t$.
2. The resistance of the resistor is $R=100ohm.$
3. Time duration from $t=0$to $t=1$ $ms.$

Joule's law of heating

1. If a conductor of resistance carries a current I for time t, then the heat developed in the conductor is proportional to the square of the current.
2. Joule's law can be defined by the form, $H=I^{2}Rt$, where, where, H is the heat developed in the conductor.

Calculation of energy dissipation

The given voltage is $E=E_o\sin \omega t$, where, $E_o=12volt$, $\omega =250\pi s^{-1}$.

So, energy dissipated in the body is

$$\begin{gathered} H=I^{2}Rt=\frac{V^2}{R}\times t(\sin ce,V=IR) \\ orH=\frac{1}{R}{\int }_0^{{10}^{-3}}{E_o}^2{\sin }^2\omega tdt. \\ orH=\frac{1}{100}{\int }_0^{{10}^{-3}}{\left(12\right)}^2{\sin }^2\omega tdt. \\ orH=\frac{144}{100}{\int }_0^{{10}^{-3}}{\sin }^2\omega tdt \\ orH=1.44{\int }_0^{{10}^{-3}}\left[\frac{1-\cos 2\omega t}{2}\right]dt \\ orH=\frac{1.44}{2}{\int }_0^{{10}^{-3}}dt-1.44{\int }_0^{{10}^{-3}}\frac{\cos 2\omega t}{2}dt \\ orH=\frac{1.44}{2}\left[t\right]{}_0^{{10}^{-3}}-\frac{1.44}{2}{\left[\frac{\sin 2\omega t}{2\omega }\right]}_0^{{10}^{-3}} \\ orH=0.72\left[{10}^{-3}-\frac{1}{2\times 250\mathrm{\pi }}\right] \\ orH=0.72\left[\frac{1}{1000}-\frac{1}{500\mathrm{\pi }}\right] \\ orH=0.72\left[\frac{\mathrm{\pi }-2}{1000}\right]=2.61\times {10}^{-4} \\ orH=2.61\times {10}^{-4}joule. \end{gathered}$$

Therefore, the energy dissipated as heat is $2.61\times {10}^{-4}joule.$