Archived Resonance and RCL Circuit

1. Oct 20, 2009

Kazk

1. The problem statement, all variables and given/known data
You have an inductor that you are planning to use in series with a variable capacitor C in the tuning section of a radio. Neglect the internal resistance of the inductor.

(a) If you have a fixed inductance L = 3.4 mH, find the maximum and minimum capacitances the variable capacitor must be able to reach in order that the resonant frequencies of the circuit cover the entire AM band: 550 - 1600 kHz.

Cmax = 2.4628e-11 F *
OK

Cmin = 2.9102e-12 F *
OK

HELP: For a given setting of the capacitance, the LC circuit has a unique resonant frequency that is given by a standard formula. (See textbook or lecture notes.)
HELP: Solve the resonant frequency expression for C as a function of the angular frequency and the inductance for the two frequencies at the ends of the AM band. Don't forget to convert frequency (in Hz) to angular frequency (in rad/s) in your numerical calculations.

(b) For the parameters of part (a), if a current is present in the circuit with peak value 2.9 µA, calculate the maximum voltage that appears across the inductor and capacitor, respectively, at the upper end of the AM frequency band.

VL max = ??????? V

VC max = ????? V

HELP: The simplest approach is to first calculate the inductive and capacitive reactances for the frequency at hand. (Consult your textbook of lecture notes for a discussion of "reactances".)
HELP: The peak voltage across any element in any AC circuit can be expressed in terms of the peak current and a resistance or reactance in a manner that is analogous to Ohm's law. (Consult your textbook or lecture notes.)

2. Relevant equations
w=2*pi*f
Xc=1/wC
Vc=IXc
XL= wL
VL=IXL

3. The attempt at a solution
w=2*pi*1600e3=1.00531e7
C used is 2.91019e-12 from previous problem.
Xc=43057.8
Vc=.124868

XL=34180.5
VL=0.099124

I have tried using the other capacitance, tried playing with frequencies, but I just can't seem to get the right answer.

2. Feb 6, 2016

Staff: Mentor

For Part (b):
When a radio is tuned to a particular frequency the reactances of the inductor and capacitor are equal at that frequency. We thus expect the peak voltage developed across the inductor and capacitor to be equal.

For the highest frequency (top of the AM band) the reactances of the inductor and capacitor are thus:

$X_{hi} = 2 \pi (1600~kHz)(3.4~mH) = 34.181~kΩ$

The peak voltage developed across the inductor and the capacitor will be the same. For a peak current of $I_p = 2.9~μA$ we have:

$V_C = V_L = I_p X_{hi} = 99.1~mV$

The OP had the correct voltage calculation for the inductor reactance and voltage. I'm not sure how he obtained the capacitive reactance he did.