Question

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

• A. $2.6\times {10}^{-3}\text{J}$
• B. $5.2\times {10}^{-3}\text{J}$
• C. $5.2\times {10}^{-4}\text{J}$
• D. $2.6\times {10}^{-4}\text{J}$

Answer 1

The correct option is D $2.6\times {10}^{-4}\text{J}$

Energy dissipated as heat at any instant of time $t$ for a very short interval $dt$ is given by,

$dE=\frac{E^2}{R}dt$

$\Rightarrow E={\int }_0^{0.001}\frac{E^2}{R}dt$

$\Rightarrow E={\int }_0^{0.001}\frac{{12}^2{\sin }^2\left(250\pi t\right)}{100}dt$

Using the formula, ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$


$E=1.44{\int }_0^{0.001}\left(\frac{1-\cos \left(500\pi t\right)}{2}\right)dt$

$\Rightarrow E=0.72{[t-\frac{\sin \left(500\pi t\right)}{500\pi }]}_0^{0.001}$

$\Rightarrow E=0.72[0.001-\frac{\sin \left(0.5\pi \right)}{500\pi }]$

$\Rightarrow E\approx 2.6\times {10}^{-4}\text{J}$

Hence, option $\left(D\right)$ is the correct answer.