# A series AC circuit containing an inductor

Question:

A series AC circuit containing an inductor $(20 \mathrm{mH})$, a capacitor $(120 \mu \mathrm{F})$ and a resistor $(60 \Omega)$ is driven by an $\mathrm{AC}$ source of $24 \mathrm{~V} / 50 \mathrm{~Hz}$. The energy dissipated in the

1. (1) $5.65 \times 10^{2} \mathrm{~J}$

2. (2) $2.26 \times 10^{3} \mathrm{~J}$

3. (3) $5.17 \times 10^{2} \mathrm{~J}$

4. (4) $3.39 \times 10^{3} \mathrm{~J}$

Correct Option: , 3

Solution:

(3)

Given: $R^{\prime}=60 \Omega, f=50 H z, \omega=2 \pi f=100 \pi$ and $v=24 v$ $\mathrm{C}=120 \mu \mathrm{f}=120 \times 10^{-6} \mathrm{f}$

$\mathrm{x}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \pi \times 120 \times 10^{-6}}=26.52 \Omega$

$\mathrm{x}_{\mathrm{L}}=\omega \mathrm{L}=100 \pi \times 20 \times 10^{-3}=2 \pi \Omega$

$\mathrm{x}_{\mathrm{C}}-\mathrm{x}_{\mathrm{L}}=20.24 \approx 20$

$z=\sqrt{R^{2}+\left(x_{C}-x_{L}\right)^{2}}$

$\mathrm{z}=20 \sqrt{10} \Omega$

$\cos \phi=\frac{\mathrm{R}}{\mathrm{z}}=\frac{60}{20 \sqrt{10}}=\frac{3}{\sqrt{10}}$

$P_{\text {avg }}=V I \cos \phi, I=\frac{V}{z}=\frac{v^{2}}{z} \cos \phi=8.64$ watt

Energy dissipated (Q) in time $\mathrm{t}=60 \mathrm{~s}$ is

$Q=P . t=8.64 \times 60=5.17 \times 10^{2} \mathrm{~J}$