Question

Comprehension :

An alternating voltage of $260\text{V}$ and $\omega =500{\text{rad s}}^{-1}$ is applied in an series $\text{LCR}$ circuit, where $L=0.01\text{H},C=4\times {10}^{-4}\text{F}$ and $R=10\Omega$

$\left(ii\right)$ Find the power supplied by the source is- (in $\text{W}$)-

• A.
  $1000$
• B.
  $6760$
• C.
  $3380$
• D.
  $3000$

Answer 1

The correct option is B

$6760$
Given, $L=0.01\text{H};C=4\times {10}^{-4}\text{F}R=10\Omega ;V_{rms}=120\text{V};\omega =500{\text{rad s}}^{-1}$

Power in an a.c. circuit is,

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

Where, $\cos \varphi =\frac{R}{Z}$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$X_L=\omega L=500\times 1\times {10}^{-2}=5$

$X_C=\frac{1}{\omega C}=\frac{1}{500\times 4\times {10}^{-4}}=5$

$\Rightarrow Z=\sqrt{(10{)}^2+(5-5{)}^2}=10\Omega$

$\Rightarrow \cos \varphi =\frac{R}{Z}=\frac{10}{10}=1$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{260}{10}=26\text{A}$

Hence, power dissipated in the circuit is,

$P=260\times 26\times 1=6760\text{W}$

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