Electrical Resonance Problem

  • Thread starter jayjay112
  • Start date
  • #1
36
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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???
 

Answers and Replies

  • #2
263
0
use the second pair of info, and the formula for the current in an RLC circuit...
 
  • #3
36
0
use the second pair of info
What do you mean by use the second pair of info?

Thanks for your help
 
  • #4
263
0
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
36
0
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??
 
  • #6
263
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yes thats cool...
 
  • #7
rl.bhat
Homework Helper
4,433
7
I = 0.707Imax when XL - XC = R
 
  • #8
36
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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 dont know V???
 
  • #9
rl.bhat
Homework Helper
4,433
7
Just calculate 2*pi*f*L -1/(2*pi*f*C). That is the value of R.
 

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