Question

Power dissipated in an LCR series circuit connected to an a.c. source of emf $\epsilon$ is :

• A. $\frac{{\epsilon }^{2}R}{\sqrt{R^2+{[L\omega -\frac{1}{C\omega }]}^2}}$
• B. $\frac{{\epsilon }^{2}R}{[R^2+{[L\omega -\frac{1}{C\omega }]}^2]}$
• C. $\frac{{\epsilon }^2\sqrt{[R^2+{[L\omega -\frac{1}{C\omega }]}^2]}}{R}$
• D. $\frac{{\epsilon }^2[R^2+{[L\omega -\frac{1}{C\omega }]}^2]}{R}$

Answer 1

The correct option is B $\frac{{\epsilon }^{2}R}{[R^2+{[L\omega -\frac{1}{C\omega }]}^2]}$

The emf of an $LCR$ circuit is $\epsilon$.

The impedance of a series $LCR$ circuit is given as,

$Z=\sqrt{\left[R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2\right]}$

The power factor in a series $LCR$ circuit is given as,

$\cos \varphi =\frac{R}{\left|Z\right|}$

The power dissipated in the circuit is given as,

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

$=\epsilon \times \frac{\epsilon }{\left|Z\right|}\times \frac{R}{\left|Z\right|}$

$=\frac{{\epsilon }^{2}R}{\left[R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2\right]}$

Thus, the power dissipated in the series $LCR$ circuit is $\frac{{\epsilon }^{2}R}{\left[R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2\right]}$.