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NCERT Class XII Chapter
Alternating Current
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Question : 10 of 26
Marks: +1, -0
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 200 μH, what must be the range of its variable capacitor?
Solution:  
For tuning, the natural frequency i.e., the frequency of L-C oscillations should be equal to frequency of radio waves received by the antenna in the form of same frequency current in the L-C circuit. For tuning at 800 kHz, required capacitance
f1f_1 = 12πLC1\frac{1}{2\pi\sqrt{LC_1}} , C1C_1 = 14π2Lf12\frac{1}{4\pi^2 L f_1^2} = 14π2(200×10−6)(800×103)2\frac{1}{4\pi^2 (200\times 10^{-6}) (800\times 10^3)^2} = 197.8 pF
For tuning of 1200 kHz, required capacitance
f2f_2 = 12πLC2\frac{1}{2\pi\sqrt{LC_2}} , C2C_2 = 14π2Lf22\frac{1}{4\pi^2 L f_2^2} = 14π2(200×10−6)(1200×103)2\frac{1}{4\pi^2 (200\times 10^{-6}) (1200\times 10^3)^2} = 87.9 pF
So, the variable capacitor should have a frequency range between 87.9 pF to 197.8 pF.
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