Question

The instantaneous values of AC and AC source in a circuit are, $i=\frac{1}{\sqrt{2}}\sin \left(100t\right)$ and $E=\frac{1}{\sqrt{2}}\sin (100t+\pi /3)$ respectively. The total work done in a cycle is -

• A. $4.85\text{mJ}$
• B. $7.85\text{mJ}$
• C. $9.85\text{mJ}$
• D. $6.85\text{mJ}$

Answer 1

The correct option is B $7.85\text{mJ}$

Given:

$i=\frac{1}{\sqrt{2}}\sin \left(100t\right)$

$E=\frac{1}{\sqrt{2}}\sin (100t+\pi /3)$

Here, the phase difference between the voltage and the current is, $\varphi =\pi /3$.

Now, average power,

$P_{\text{avg}}=E_{\text{rms}}{i}_{\text{rms}}\cos \varphi$

$\Rightarrow P_{\text{avg}}=\frac{E_0}{\sqrt{2}}\frac{i_0}{\sqrt{2}}\cos \varphi$

$\Rightarrow P_{\text{avg}}=\frac{1/\sqrt{2}}{\sqrt{2}}\times \frac{1/\sqrt{2}}{\sqrt{2}}\times \cos \left(\pi /3\right)=\frac{1}{8}\text{W}$

Now, the total work done in a cycle is,

$W=P_{\text{avg}}T=P_{\text{avg}}\times \frac{2\pi }{\omega }=\frac{1}{8}\times \frac{2\pi }{100}=7.85\times {10}^{-3}\text{J}=7.85\text{mJ}$

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