Inductor and Variable Capacitor?

Question

(http://i68.photobucket.com/albums/i19/r34racer01/showmepl2-9.gif?t=1270572801)

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

(a) If you have a fixed inductance L = 2.5 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. Neglect the internal resistance of the inductor.
Cmax = ? F
Cmin = ? F
(b) For the parameters of part (a), if a current is present in the circuit with peak value 3 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
(c) Suppose now you wish to build a variable LC circuit whose resonant frequencies cover the full FM band: 88 - 108 MHz. If you choose a fixed inductance L = 29 µH, what are the maximum and minimum capacitances the variable capacitor must reach in this case?
C'max = ? F
C'min = ? F

Answer 1

A. First we put the frequencies into rad/s.

w1 = 5.5E5*2pi = 3455751.91894877 rad/s

w2 = 1.6E6*2pi = 10053096.4914873 rad/s

(BTW, I'm not showing off with all these sig. figs, it's just easier to copy/paste straight from my calculator.)

Then we solve for C: w = 1/sqrt(LC) ==> C = 1/(w^2L)

C1 = 3.34946061627564D-11 F

C2 = 3.95785873602883D-12 F

B. We know that at resonance, XL = -XC, so the peak V and I are the same in the capacitor and the inductor. Arbitrarily choosing XL to work with,

V = I*XL (Note, these are not Roman numerals; nor is C!)

XL = wL

VL max = VC max = I*XL = I*w2*L = 0.075398223686155 V

C. Just repeat step A with the two FM frequencies and the new value for L.



I hope you're starting to get the hang of these problems. Given the questions you've already asked and gotten answers to, this one should have been relatively easy. A quick review:

XL = wL

XC = -1/(wC)

X = XL+XC

At resonance (as in a tuner), XL = -XC which leads to

(1) Resonant frequency w = 1/sqrt(LC) rad/s

(2) X = 0

At any frequency, the reactive voltage Vx = IX, just like the resistive voltage Vr = IR. The difference is that X changes with frequency and R doesn't. Also, in an RLC circuit not at resonance, Vx and Vr differ in phase by 90 deg or pi/2 rad. The total voltage V = sqrt(Vx^2+Vr^2).

In a series-resonant LC circuit Vx is the algebraic sum of V(L) and V(C). V(L) and V(C) may be, and at resonance are, much larger than Vx, but they are phase-opposed and thus sum to a smaller value.

In a parallel-resonant LC circuit, the same statements can be made about the currents Ix, I(L) and I(C).

I hope this will help you have more confidence in, and place more reliance on, your own work. Email me if you want to discuss this further.

Answer 2

If I remember correctly, an LC filter contains an Inductor (L) and Capacitor(C). On schematic diagrams, Inductors are normally labeled L. The bleeder resistor gets rid of any built up charge left in the capacitor after the input power has been removed. When you cut the input power and the current stops, the inductor's magnetic field collaspes. This collaspe induces the capcitor to charge up again. The bleeder resistor gets rid of this charge. I'm pretty sure...

Answer 3

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Answer 4

You should be able to pick up a classdic radio station