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A radio cn tune over the frequency range of a portion of Mw broadcast band (800 kHz to 1200 kHz ) . If its circuit has an effective inductance of 200 muH, what must be the range of its veriable capacitor ? |
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Answer» Solution :Frequency of OSCILLATION is given by f = `(1)/(2 pi) (1)/(sqrt(LC))` Where L is the inductance and C is the capacitance of oscillating circuit `THEREFORE "" C = (1)/(4 pi^(2)f^(2) L)` Now For `L_(1) = 200 mu H= 2 xx 10^(-4) `H and `f_(1) = 1200` kHz = 12 `xx 10^(5)` HZ `C_(1) = (1)/(4 xx pi^(2) xx (12 xx 10^(5))^(2) xx 2 xx 10^(-4))` `rArr "" C_(1) = ` 88.04 pF for `"" L_(2) = 200 mu H = 2 xx 10^(-4) ` H `f_(2)= 800 ` kHz = `8 xx 10^(5)` Hz `C_(2) = (1)/( 4 xx pi^(2) (8 xx 10^(5))^(2) xx 2 xx 10^(-4))` `rArr "" C_(2) = `197 .8 pF The variable capacitor should have a RANGE of about 8 pF to 198 pF. |
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