Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A radio can tune over the frequency range of a portion of MW broadcast band:

Q7.10  A radio can tune over the frequency range of a portion of MW broadcast band:  (800kHz\: to1200kHz)  . If its LCcircuit has an effective inductance of 200\mu H, what must be the range of its variable capacitor?

Answers (1)

best_answer

Given, 

Range of the frequency in which radio can be tune = (800kHz\: to1200kHz)

The effective inductance of the Circuit = 200\mu H

Now, As we know,

 w^2=1/\sqrt{LC}

C=1/w^2L

where w is tuning frequency.

For getting the range of the value of a capacitor, let's calculate the two values of the capacitor, one maximum, and one minimum.

first, let's calculate the minimum value of capacitance which is the case when tuning frequency = 800KHz.

C_{minimum}=\frac{1}{w_{minimum}^2L}=\frac{1}{(2\pi(800*10^3))^2*200*10^{-6}}=1.981*10^{-10}F

Hence the minimum value of capacitance is 198pF.

Now, Let's calculate the maximum value of the capacitor.

in this case, tuning frequency = 1200KHz

C_{maximum}=\frac{1}{w_{maximum}^2L}=\frac{1}{(2\pi(1200*10^3))^2*200*10^{-6}}=88.04*10^{-12}F

Hence the maximum  value of the capacitor is 88.04pf

Hence the Range of the values of the capacitor is 88.04pF to 198.1pF.

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