Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Figure shows a series LCR circuit connected to a variable frequency 230 V source. Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

Q7.11 (b) Figure  shows a series LCR circuit connected to a variable frequency 230\: V source.

                L=5.0H,  C=80\mu F, R=40\Omega.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

            

Answers (1)

best_answer

Given,

Variable frequency supply voltage V = 230V

Inductance L=5.0H

Capacitance C=80\mu F=80*10^{-6}F

Resistance R=40\Omega

Now,

The impedance of the circuit is 

Z=\sqrt{(wL-\frac{1}{wC})^2+R^2} 

at Resonance Condition

:wL=\frac{1}{wC}

Z=R=40\Omega

Hence, Impedance at resonance is 40\Omega.

Now, at resonance condition, impedance is minimum which means current is maximum which will happen when we have a peak voltage, so

Current in the Resonance circuit is Given by 

I_{resonance}=\frac{V_{peak}}{Z}=\frac{\sqrt{2}* 230}{40}=8.13A

Hence amplitude of the current at resonance is 8.13A.

 

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 board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question