# Electrical Resonance Problem

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 I am 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?

## Answers and Replies

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

jayjay112
use the second pair of info

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

Thanks for your help

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...

jayjay112
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...

oh thanks,

so is the first part correct??

Thaakisfox
yes that's cool...

Homework Helper
I = 0.707Imax when XL - XC = R

jayjay112
I = 0.707Imax when XL - XC = R

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 don't know V?

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