Question

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

• A. $5.65\times {10}^2\text{J}$
• B. $2.26\times {10}^3\text{J}$
• C. $5.17\times {10}^2\text{J}$
• D. $3.39\times {10}^3\text{J}$

Answer 1

The correct option is C $5.17\times {10}^2\text{J}$

Given:
$R=60\Omega ,f=50\text{Hz},\omega =2\pi f=100\pi \text{rad/s}$ and $V=24\text{V}$

$C=120\mu \text{F}=120\times {10}^{-6}\text{F}$

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

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

$X_C-X_L=26.52-2\pi =20.24\approx 20\Omega$


Now, $Z=\sqrt{R^2+(X_C-X_L{)}^2}=\sqrt{{60}^2+(20{)}^2}=20\sqrt{10}\Omega$

So, $\cos \varphi =\frac{R}{Z}=\frac{60}{20\sqrt{10}}=\frac{3}{\sqrt{10}}$

As power, $P_{avg}=VI\cos \varphi$ and $I=\frac{V}{Z}$

$\therefore P_{avg}=\frac{V^2}{Z}\cos \varphi =\frac{{24}^2}{20\sqrt{10}}\times \frac{3}{\sqrt{1}0}=8.64\text{W}$

Energy dissipated $\left(Q\right)$ in time $t=60\text{s}$ is,

$Q=P_{avg}.t=8.64\times 60=5.17\times {10}^2\text{J}$

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