Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.10
NCERT Solutions for Class 12 Physics Chapter 7 Question 10
A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200μH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.]
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The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.
Lower tuning frequency, ν1 = 800 kHz = 800 x 103 Hz
Upper tuning frequency, ν2 = 1200 kHz = 1200 x 103 Hz
Effective inductance of circuit L = 200μH = 200 x 10⁻6 H
Capacitance of variable capacitor for ν1 is given as:
C₁ = 1/(ω1)²L
Where, ω1 = Angular frequency for capacitor C₁= 2πν1= 2π x 800 x 103 rad/s
Therefore,
C₁ = 1/(2π x 800 x 10³ )² x 200 x 10⁻6 = 1.9809 x 10⁻¹⁰ F = 198 pF
Capacitance of variable capacitor for ν2 is given as ;
C2 = 1/(ω2)²L
Where, ω2 = Angular frequency for capacitor C2 = 2πν2 = 2π x 1200 x 103 rad/s
Therefore ,
C2 = 1/(2π x 1200 x 10³ )² x 200 x 10⁻6 = 0.8804 x 10⁻¹⁰ F = 88 pF
Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.