Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q7.20 (d) A series LCR circuit with $L=0.12H$ , $C=480nF$ , $R=23\Omega$ is connected to a $230V$ variable frequency supply. For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

Answers (1)

best_answer

As

Power $P=I^2R$

Power $P$ will be half when the current $I$ is $1/\sqrt{2}$ times the maximum current.

As,

$I =I_{max}Sin(wt-\phi)$

At half power point :

$\frac{i_{max}}{\sqrt{2}} =I_{max}Sin(wt-\phi)$

$\frac{1}{\sqrt{2}} =Sin(wt-\phi)$

$wt=\phi+\frac{\pi}{4}$

here,

$\phi=tan^{-1}(\frac{wL-\frac{1}{wC}}{R})$

On putting values, we get, two values of $w$ for which

$wt=\phi+\frac{\pi}{4}$

And they are:

$w_1=678.75Hz$

$w_2=648.22Hz$

Also,

The current amplitude at these frequencies

$I_{halfpowerpoint}=\frac{I_{max}}{\sqrt{2}}=\frac{14.14}{1.414}=10A$

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

Latest Question