Question

A circuit connected to an $ac$ source of $\text{emf}e=e_0\sin \left(100t\right)$ with $t$ in $\text{seconds}$, gives a phase difference of $\frac{\pi }{4}$ between the $\text{emf}e$ and current $i$. Which of the following circuits will exhibit this?

• A. RL circuit with $R=1\text{k}\Omega$ and $L=1\text{mH}$
• B. RL circuit with $R=1\text{k}\Omega$ and $L=10\text{mH}$
• C. RC circuit with $R=1\text{k}\Omega$ and $C=10\mu \text{F}$
• D. RC circuit with $R=1\text{k}\Omega$ and $C=1\mu \text{F}$

Answer 1

The correct option is C RC circuit with $R=1\text{k}\Omega$ and $C=10\mu \text{F}$

No information is given about leading or lagging quantity between emf and current. So, the circuit can be RC or RL.

Now, $\omega =100\text{rad/s}$
As,
$\tan \varphi =\frac{X}{R}$

Here, $X$ can be either $X_C$ or $X_L$

Considering $R=1k\Omega$, as this value is common to all options.

So, $\tan {45}^{\circ }=\frac{X}{R}\Rightarrow R=X$

$\therefore X=1000\Omega$

Now, considering $X_L=1000\Omega$

$\Rightarrow \omega L=1000$

$\Rightarrow 100\times L=1000\Rightarrow L=10\text{H}$

Now, Considering $X_C=1000\Omega$

So, $\frac{1}{\omega C}=1000$

$\frac{1}{100\times C}=1000\Rightarrow C={10}^{-5}\text{F}=10\mu F$

So, the phase difference between emf and current will be ${45}^{\circ }$ if

$R=1k\Omega ,L=10\text{H}$ or
$R=1k\Omega ,C=10\mu F$

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