Question

For the given circuit, the average power developed is -

• A. $50\sqrt{2}\text{W}$
• B. $200\text{W}$
• C. $50\sqrt{2}\text{W}$
• D. $200\sqrt{2}\text{W}$

Answer 1

The correct option is B $200\text{W}$

Given,

$V=200\sin \left(250t\right)\text{V}L=0.2\text{H};R=50\Omega$

The average power of an a.c. circuit is,

$P=V_{rms}{I}_{rms}\cos \varphi$

$P=V_{rms}\times \frac{V_{rms}}{Z}\times \frac{R}{Z}={\left(\frac{V_{rms}}{Z}\right)}^{2}R$

$\because$ Impedance of $RL$ circuit is given by

$Z=\sqrt{R^2+(\omega L{)}^2}=\sqrt{(50{)}^2+(0.2\times 250{)}^2}$

$=\sqrt{(50{)}^2+(50{)}^2}=50\sqrt{2}\Omega$

Now, the average power is,

$P={\left(\frac{200}{\sqrt{2}}\right)}^2\times \frac{50}{50\sqrt{2}}\times \frac{1}{50\sqrt{2}}=200\text{W}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.