Question

A series RCL circuit has a resonant frequency of 1470 Hz. When operating at a frequency other than 1470 Hz, the circuit has a capacitive reactance of 5.0 ? and an inductive reactance of 25.0 ?.

(a) What is the value of L?

(b) What is the value of C?

Me and a friend have been trying to figure this out for a while. I know I have the XL = 2*pi*f*L formula.

Any help would be appreciated (particularly if its in detailed/clear steps) so I can figure out how to do this problem that could very well be on the test.

Answer 1

a.) The formula for capacitive reactance, Xc = 1 / [2(Pi) f C], where
f is the frequency and C is the capacitance of the capacitor

The formula for inductive reactance, Xl = 2(Pi) f L, where
L is the inductance of the inductor

In a series RCL circuit resonance occurs when Xc = Xl.
Setting these two equations equal you get:

1 / [2(Pi) f C] = 2(Pi) f L

4[(Pi)^2] (f^2) LC = 1

f^2 = 1 / [4[(Pi)^2] LC]
f = 1 / [2(Pi) SQRT(LC)]
Since the resonant frequency is given as 1470 Hz:
1470 = 1 / [2(Pi) SQRT(LC)]
LC = 1.17 x 10^-8

Xc = 1 / [2(Pi) f C] = 5.0 ohms
1 / [2(Pi) f C] = 5.0
f C = 0.0318

Xl = 2(Pi) f L = 25.0 ohms
2(Pi) f L = 25.0
f L = 3.978

Dividing f C = 0.0318 by f L = 3.978, you get:
C / L = 0.0318 / 3.978 = 0.00799
C = 0.00799 L

Combining this result with the result from resonance analysis,
LC = 1.17 x 10^-8, you get:
L(0.00799 L) = 1.17 x 10^-8
L^2 = 1.46 x 10^-6
L = 0.001208H

b.) since LC = 1.17 x 10^-8, and L = 0.001208H
C = 9.68 x 10^-6 F