Question

An $AC$ voltage $E=E_0\sin 100t$ is applied to the circuit where all the elements are in series, the phase difference between the current and the voltage is found to be $\frac{\pi }{4}$, then

• A. The given circuit is $LR$ circuit.
• B. The given circuit is $RC$ circuit.
• C. $R=1\text{k}\Omega$ and $C=10\mu \text{F}$
• D. $L=1\text{H}$ and $R=10\Omega$

Answer 1

Answer: (B), (C)

$R=1\text{k}\Omega$ and $C=10\mu \text{F}$

Given,

Phase difference between voltage $V$ and current $I$ is $\frac{\pi }{4}$ and the current is leading here.

As, the current leads voltage only in capacitive circuit, so, the above graph is drawn for $CR$ circuit.

Thus, we can eliminate the options $A$ and $D$.

Assuming the circuit to be $RC$ circuit

$\therefore \tan \varphi =\frac{X_C}{R}$

$\Rightarrow \frac{X_C}{R}=\tan \frac{\pi }{4}=1$

$\Rightarrow R=X_C=\frac{1}{\omega C}$

$\Rightarrow R=\frac{1}{100C}$

If, $R=1\text{k}\Omega \Rightarrow C=10\mu \text{F}$

Therefore, option $\left(C\right)$ satisfies the above equation.

So, the given circuit is $RC$ circuit.

Hence, options $\left(B\right)$ and $\left(C\right)$ are correct.