Question

A circuit, containing an inductance and a resistance connected in series, has an AC source of $200\text{V},50\text{Hz}$ connected across it. An AC rms current of $10\text{A}$ flows through the circuit and the power loss is measured to be $1\text{kW}$. Then

• A. The inductance of the circuit is $\frac{\sqrt{3}}{10\pi }\text{H}$.
• B. The frequency of the AC when the phase difference between the current and emf becomes $\frac{\pi }{4}$, with the above components is $\frac{50}{\sqrt{3}}\text{Hz}$
• C. The frequency of AC, at which the reactive power is half of the actual power lost is $\frac{25}{\sqrt{3}}\text{Hz}$
• D. The frequency of the AC when the phase difference between the current and emf becomes $\frac{\pi }{4}$, with the above components is $\frac{25}{\sqrt{3}}\text{Hz}$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The inductance of the circuit is $\frac{\sqrt{3}}{10\pi }\text{H}$.
B The frequency of the AC when the phase difference between the current and emf becomes $\frac{\pi }{4}$, with the above components is $\frac{50}{\sqrt{3}}\text{Hz}$
C The frequency of AC, at which the reactive power is half of the actual power lost is $\frac{25}{\sqrt{3}}\text{Hz}$


$\left(A\right)$
The power loss in the circuit is
$P_{loss}=V_{rms}\times I_{rms}\cos \varphi =200\times 10\times \cos \varphi =1000\text{W}$
$\cos \varphi =\frac{1}{2}$
$\therefore \varphi =\frac{\pi }{3}$
(since, the circuit element is an inductor)
The inductance in the circuit is calculated as follows:
$X_L=\frac{200}{10}\times \sin \frac{\pi }{3}=10\sqrt{3}$
$\therefore L=\frac{10\sqrt{3}}{2\pi \times 50}=\frac{\sqrt{3}}{10\pi }\text{H}$


$\left(B\right)$
$I=\frac{V}{Z}$
$10=\frac{200}{\sqrt{X_L^2+R^2}}$
$400=X_L^2+R^2,\because X_L=\sqrt{3}R$
$4R^2=400$
$\Rightarrow R=10\Omega$

$\frac{\omega L}{R}=\tan \frac{\pi }{4}=1$
$2\pi f\times \frac{\sqrt{3}}{10\pi }\times \frac{1}{10}=1$
$f=\frac{50}{\sqrt{3}}\text{Hz}$.

$\left(C\right)\frac{P_{\text{reactive}}}{P_{\text{loss}}}=\frac{1}{2}=\tan \varphi =\frac{\omega L}{R}$
$\therefore f=\frac{25}{\sqrt{3}}\text{Hz}$,
which is half of the previous value.