Question

An a.c generator supplying voltage $E=10\sin \omega t\text{V}$ to a circuit which consists of a resistor of resistance $3\Omega$ and an inductor of reactance $4\Omega$ as shown in the figure. The voltage across the inductor at $t=\frac{\pi }{\omega }$ is

• A. $2\text{V}$
• B. $10\text{V}$
• C. Zero
• D. $4.8\text{V}$

Answer 1

The correct option is D $4.8\text{V}$

Impedance of the circuit $Z=\sqrt{R^2+X_L^2}$
$=\sqrt{3^2+4^2}=5\Omega$

Power factor of the circuit $\cos \varphi =\frac{R}{Z}=\frac{3}{5}$
$\Rightarrow \varphi ={53}^{\circ }$

Therefore, current in the circuit is
$\because i=\frac{E_0}{Z}\sin (\omega t-{53}^{\circ })=2\sin (\omega t-{53}^{\circ })$

Voltage across the inductor
$V_L=L\frac{di}{dt}=2\omega L\cos (\omega t-{53}^{\circ })$
At $t=\frac{\pi }{\omega }$, $V_L=8\cos (\pi -{53}^{\circ })$
($\because \omega L=4$)
$V_L=-8\cos {53}^{\circ }=-8\times \frac{3}{5}$
$\therefore V_L=-4.8\text{V}$
i.e. potential across inductor lags behind.