# LC circuit with variable capacitor

1. Aug 28, 2011

### Piano man

1. The problem statement, all variables and given/known data
A constant voltage at a frequency of 1MHz is maintained across a circuit consisting of an inductor in series with a variable capacitor. When the capacitor is reduced from 300pF to 284pF, the current is 0.707 of its maximum value. Find
(i) the inductance and the resistance of the inductor, and
(ii) the Q factor of the inductor at 1MHz. Sketch the phasor diagram for each condition.

2. Relevant equations
$$\omega=\frac{1}{\sqrt{LC}} // v_L=L\frac{di}{dt} // v_C=\frac{q}{C}$$
Probably a load of other formulae as well.

3. The attempt at a solution

Using Kirchoff, get $$L\frac{di}{dt} + \frac{q}{300*10^{-12}}=0$$ for the first case and $$0.707L\frac{di}{dt}+\frac{0.707it}{284*10^{-12}}=0$$ for the second.
Equating the two, I'm still left with i and t and a feeling that I'm barking up the wrong tree.
Any help would be appreciated - it's ages since I've done these type of questions.
Thanks.

2. Aug 28, 2011

### Spinnor

Do you take into account the voltage generated by the resistance of the inductor?

Last edited: Aug 28, 2011
3. Sep 1, 2011

### rude man

Unfortunately, this problem as stated gives insufficient info.

As a matter of fact, one solution is Q = infinity, L = (√2/wC1 - 1/wC2)/w(√2-1) H

where C1 = 300 pF, C2 = 284pF, R = 0 and w = 2pi*1 MHz.

As Spinnor indicates, the same conditions can obtain when R (and therefore Q) are finite. What has to be specified in addition to the above is the phase of the current i0/√2. I happen to have picked 180 deg. I suppose one could object that I went from 0 to 180 which is -0.707i0 but remember if the detuning is to be ascribed to a finite Q, i0/√2 is not in phase with i0 either .... out 45 deg in fact.