You have an inductor that you are planning to use in series witha variable capac

Question

You have an inductor that you are planning to use in series witha variable capacitor C in the tuning section of aradio.

(a) If you have a fixed inductance L = 3.1 mH, find themaximum and minimum capacitances the variable capacitor must beable to reach in order that the resonant frequencies of the circuitcover the entire AM band: 550 - 1600 kHz. Neglect the internalresistance of the inductor.

Cmax = F   

Cmin = F    

(b) For the parameters of part (a), if a current is present inthe circuit with peak value 2.5 µA, calculate the maximumvoltage that appears across the inductor and capacitor,respectively, at the upper end of the AM frequency band.

VL max = V   

VC max = V    

(c) Suppose now you wish to build a variable LC circuit whoseresonant frequencies cover the full FM band: 88 - 108 MHz. If youchoose a fixed inductance L = 32 µH, what are themaximum and minimum capacitances the variable capacitor must reachin this case?

C'max = F   

C'min = F

Explanation / Answer

Part a: f_max = 1/(2*LC) f_max = 1600*10^3 L = 3.1*10^-3 = 0.0031 LC = 1/(2f_max) 1/2f = 9.95*10^-8 LC = 9.95*10^-8 Squaring both sides LC = 9.8947*10^-15 Cmax = (9.8947*10^-15)/0.0031 = 3.198*10^-12 F For Cmin apply the lower frequency 550KHZ 1/2f = 2.894*10^-7 Cmin = 8.374*10^-14/0.0031 = 2.7012*10^-11 F Part b: Ipeak = 2.5*10^-6 A Inductive Reactance = 2*(f_max)*L =2*1600*10^3*3.1*10^-3 = 31164.6 Voltage across the inductor = 2.5*10^-6*31164.6 = 0.07791 V Capacitive Reactance = 1/(2f*C) =1/(2*1600*10^3*3.198*10^-12) = 31104.4 Votage across the capacitor = 2.5*10^-6*31104.4 = 0.077761 V Part c: L = 32*10^-6 fmax = 108*10^6 fmin = 88*10^6 After inputing the values you get the following answers Cmax = 6.7865*10^-14F Cmin = 10.222*10^-13F

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