Ncert Exercises & Solutions

Question 7.10:

A radio can tune over the frequency range of a portion of the MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 µH, what must be the range of its variable capacitor?

[ Hint: The condition for tuning is that the natural frequency which is the frequency of free oscillations of the LC network must be of the same value as the radio wave frequency ]

The range of frequency (v) of radio is 800 kHz to 1200 kHz.
Lower tuning frequency, vi = 800 kHz = 800x 10³ Hz
Upper tuning frequency, v2 = 1200 kHz = 1200 x 10³ Hz
Effective inductance of circuit L = 200 $μ$ H = 200 x $10^{- 6}$ H
The capacitance of the variable capacitor for vi is given as:

$à C_{1} = \frac{1}{ω_{1}^{2} L}$

Where, $ω_{1}$ = Angular frequency for capacitor $C_{1} = 2 {πv}_{1} = 2 π x 800 x 10^{3} rad / s$

$∴ C_{1} = \frac{1}{{(2 π \times 800 \times 10^{3})}^{2} \times 200 \times 10^{- 6}} = 1.9809 \times 10^{- 10} F = 198 pF$

The capacitance of the variable capacitor for vz is given as:

$C_{2} = \frac{1}{ω_{2}^{2} L}$

Where, $ω_{2}$ = Angular frequency for capacitor $C_{2} = 2 {πv}_{2} = 2 π x 1200 x 10^{3} rad / s$

$∴ C_{2} = \frac{1}{{(2 π \times 1200 \times 10^{3})}^{2} \times 200 \times 10^{- 6}} = 0.8804 \times 10^{- 10} F = 88 pF$

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

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