Question

An ac-generator supplying voltage $E$ 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 $t=\pi /\omega$ is

• A. $2\text{volts}$
• B. $10\text{volts}$
• C. zero
• D. $4.8\text{volts}$

Answer 1

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

Given:
$R=3\Omega ,X_L=4\Omega ,E=10\sin \omega t,t=\frac{\pi }{\omega }\text{sec}$.

Impedance of $\text{L-R}$ circuit is , $Z=\sqrt{{\left(R\right)}^2+{\left(X_L\right)}^2}=\sqrt{3^2+4^2}=5\Omega$

and , $I_0=\frac{E_0}{Z}=\frac{10}{5}=2\text{A}$

Phase angle in $\text{L-R}$ circuit is , $\theta ={\tan }^{-1}\left(\frac{X_L}{R}\right)={\tan }^{-1}\left(\frac{4}{3}\right)\Rightarrow \theta ={53}^{\circ }$

As we know that in inductor voltage leads to current by a phase of $\frac{\pi }{2}$.

$\therefore$ voltage and current can be written in $\text{L-R}$ circuit as follows.

$i=2\sin (\omega t-{53}^{\circ })$ and $V_L=8\sin (\omega t-{53}^{\circ }+{90}^{\circ })$

$\Rightarrow V_L=8\cos (\omega t-{53}^{\circ })$

$V_L=8\cos (\omega t-{53}^{\circ })$

$V_L=|8\cos {53}^{\circ }|(\text{at}t=\frac{\pi }{\omega })$

$\Rightarrow V_L=4.8\text{volts}$

Hence, option (d) is the correct answer.