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 $60\Omega$ is driven by an $\text{AC}$ source of $24\text{V},50\text{Hz}$. The energy dissipated in the circuit in $60\text{s}$ is

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

Answer 1

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



Given:

$R=60\Omega ,f=50\text{Hz},\omega =2\pi f=100\pi V=24\text{V},C=120\mu \text{F}=120\times {10}^{-6}\text{F}$
Capacitative reactance si
$X_C=\frac{1}{\omega C}=\frac{1}{100\pi \times 120\times {10}^{-6}}=26.52\Omega$
Inductive reactance is
$X_L=\omega L=100\pi \times 20\times {10}^{-3}=2\pi \Omega$
$X_C-X_L=20.24\approx 20$

Impedance is

$Z=\sqrt{R^2+(X_c-X_L{)}^2}$
$Z=20\sqrt{10}\Omega$
$\cos \varphi =\frac{R}{Z}=\frac{60}{20\sqrt{10}}=\frac{3}{\sqrt{10}}$

Avg power consumed in AC circuit is
$P_{avg}=VI\cos \varphi ,(\because I=\frac{V}{Z})P_{avg}=\frac{V^2}{Z}\cos \varphi =8.64\text{Watt}$
Energy dissipated $\left(Q\right)$ in time $t=60\text{s}$ is
$Q=P.t=8.64\times 60=5.17\times {10}^2\text{J}$