Solve RLC Circuit Power Factor Problem: 3 x 10^3 Hz, 139V rms, 37.1A rms

[yankeekd25]

Homework Statement

When the power factor of RLC circuit is equal to one, the frequency of the voltage source is 3 x 10^3 Hz. The rms value of the voltage source is 139 Volts and at a frequency of 3 x 10^3 Hz, the rms current in the circuit is 37.1 amps. If the inductive reactance at 3 x 10^3 Hz is 47 Ohms, what is the average power of the circuit in Watts at 0.78 times the resonant frequency of the circuit?

Homework Equations

Pav= Irms Vrms cos theta
Irms= I/ rad 2
Vrms= V/rad 2

The Attempt at a Solution

I have no idea where to begin. I keep going in circles. The only thing I know is cos theta=1.
Can someone please help me out with this one?

 

[gneill]

A complete solution is offered.

We assume a series RLC circuit since it wasn't specified and no diagram was included.

When the power factor is one the current is in phase with the voltage, the net reactance is zero, and the circuit is at resonance. Thus we are given the following facts for resonance conditions:
$f_o = 3~kHz$
$V_s = 139~V$ (rms)
$I = 37.1 A$ (rms)
$X_L = 47~Ω$

We can compute the inductance:
$L = \frac{X_L}{2 \pi f_o} = 2.493~mH$

At resonance the capacitive reactance will equal the inductive reactance, so:
$C = \frac{1}{2 \pi f_o X_L} = 1.129~μF$

The new operating frequency of the circuit is at $f = 0.78 f_o = 2.340~kHz$. We'll use impedances to find the current and power under the new conditions. Note that the resistance doesn't change with frequency.

$Z_L = j 2 \pi f L = j36.660 Ω$
$Z_C = \frac{1}{j 2\pi f C} = -j60.256~Ω$

The net impedance of the series circuit is:
$Z = R + Z_L + Z_C = 3.747 - j23.596~Ω$

The current and power:
$I = \frac{V_s}{Z} = 0.912 + j5.746~A~~~$ or $~~~I = 5.818~A ~~ ∠80.98°$
$P = E I cos(Φ) = 126.8~W$