# Electrical Resonance Problem

1. Apr 10, 2009

### jayjay112

A constant voltage at a frequency of 1MHz is maintained across a circut consisting of the inductor in series with a variable capacitor. When the capacitor is set to 300pF the current has its max value. When the capacitor is reduced to 284pF the current is 0.707 of its ma value. Find

The inductance and the resistance of the inductor

Im not sure if im using the correct formula.

f=[1/(2.pi)] x [1/(square root of (L).(C)]

f=frequency
L=Inductance
C=capacitance

L = 1/[(300x10^-12) x (2,000,000.pi)^2]

=337micro H

What formula could I use to find the resistance of the inductor???

2. Apr 10, 2009

### Thaakisfox

use the second pair of info, and the formula for the current in an RLC circuit...

3. Apr 10, 2009

### jayjay112

What do you mean by use the second pair of info?

4. Apr 10, 2009

### Thaakisfox

well you know that when the capacity is 284pF then the current is 0.707 times the max...

So use the general formula for the current (containing R) and make it equal to 0.707 times the max current...

5. Apr 10, 2009

### jayjay112

oh thanks,

so is the first part correct??

6. Apr 10, 2009

### Thaakisfox

yes thats cool...

7. Apr 10, 2009

### rl.bhat

I = 0.707Imax when XL - XC = R

8. Apr 10, 2009

### jayjay112

So i use this formula

I = V.R.C/L

I(max) = [V. R. (300x10^-12)] / (337x10^-6)

and then I = 0.707 I(max)

But i dont know V???

9. Apr 10, 2009

### rl.bhat

Just calculate 2*pi*f*L -1/(2*pi*f*C). That is the value of R.