Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q7.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply $(240 V\: ,10 kHz)$ . Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Answers (1)

best_answer

Given,

The inductance of the coil $L=50H$

the resistance of the coil $R=100\Omega$

Supply voltage $V=240V$

Supply voltage frequency $f=10kHz$

a)

Now, as we know peak voltage = $\sqrt2$ (RMS Voltage)

Peak voltage $V_{peak}=\sqrt2*240=339.4V$

Now,

The impedance of the circuit :

$Z=\sqrt{R^2+(wL)^2}=\sqrt{100^2+(2\pi 10*10^3 *50)^2}$

Now peak current in the circuit :

$I_{peak}=\frac{V_{peak}}{Z}=\frac{339}{\sqrt{100^2+(2\pi 10*10^3 *50)^2}}=1.1*10^{-2}A$

Hence peak current is $1.1*10^{-2}A$ in the circuit.

The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit in the case of high frequency.

b)

For phase difference $\phi$ we have

$tan\phi =\frac{wL}{R}=\frac{2\pi *10*10^3*0.5}{100}=100\pi$

$\phi =89.82^0$

Now

$t=\frac{\phi}{w}=\frac{89.82*\pi}{180*2\pi *10^3}=25\mu s$

Hence time lag between the maximum voltage and the maximum current is $25\mu s$ .

In the DC circuit, after attaining the steady state, inductor behaves line short circuit as $w$ is Zero.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th

anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question