# Archived Power factor at half power points for RLC circuit

1. May 12, 2004

### Loki

Can anyone help me with this homework problem? I worked it through, but the answer i get doesn't agree with the answer the book gives. I'll include my work, and if anyone can help, I'll be eternally grateful, at least for a while.

The Problem:

In a series RCL circuit the dissapated power drops by a factor of two when the frequency of the generator is changed from the resonant frequency to a nonresonant frequency. The peak voltage is held constant while this change is made. Determine the power factor of the curcuit at nonresonant frequency.​

My Work:

$$P_r = 2 P_{nr}$$
$$P = I * V * \cos \phi$$
$$\phi = \arctan \frac{X_L - X_C}{R}$$
At resonant frequency, $X_L - X_C = 0$
$$I * V * \cos ( \arctan \frac{X_L-X_C}{R} ) = 2 * I * V \cos (\arctan\frac{X_L-X_C}{R})$$
$$I * V * \cos ( \arctan (0)) = 2 * I * V \cos (\arctan\frac{X_L-X_C}{R})$$
$$I * V * \cos (0) = 2 * I * V \cos (\phi)$$
$$1 = 2 \cos (\phi)$$
$$\cos \phi = \frac{1}{2}$$

This seems correct to me, but the book says the answer is $\frac{\sqrt {2}}{2}$

Am I mistaking the "Factor of two" part of the problem? Does this really mean that the power is squared?

If anybody can help, thanks. On a sidenote, LaTeX is very nice, if a bit tedious to write in.

Last edited by a moderator: May 14, 2004
2. Feb 9, 2016

### Staff: Mentor

A complete solution is offered.

At resonance the current is in phase with the voltage source, and the power factor is 1. In that case the power dissipated is simply

$P_o = \frac{V^2}{R}~~~$where $R$ is the circuit resistance and $V$ the fixed voltage.

We are looking for the power factor for the situation where the dissipated power drops by a factor of two. That is,

$P = \frac{P_o}{2} = \frac{V^2}{2R}$

Off-resonance there will be a net reactance, X. We won't worry about the individual reactances of the inductance and capacitance, only their net effect. Let:

$Z = R + j X$ be the impedance at the off-resonance frequency.

Then the complex power (apparent power) for the circuit is given by:

$p = V~I^* = V ⋅ \frac{V}{(R + jX)^*} = V^2 \frac{R + jX}{R^2 +X^2}$

where $I^*$ means the complex conjugate of $I$.

Now, the real power dissipated is the real component of the complex power. Setting this equal to our desired value:

$\frac{V^2}{2R} = V^2 \frac{R}{R^2 +X^2}$

Solving for the reactance we find that $X = R$. This means that the impedance has a value $Z = R (1 + j)$, which has a phase angle of 45°. The current, which is V/Z will have an angle of -45°. The cosine of -45° is $\frac{\sqrt{2}}{2}$, so that is the power factor.

Note that the reactance could be positive or negative (net inductive or net capacitive reactance), and that the power factor will be the same since $~~~cos(-45°) = cos(45°) = \frac{\sqrt{2}}{2}$.

3. Feb 9, 2016

### cnh1995

Here's my two cents..
At resonance,
Pr=VIr..(cosΦ=1)
At non-resonant frequency,
Pnr=VInrcosΦ
But Pr=2Pnr
So,
VIr=2VInrcosΦ
∴Ir=2InrcosΦ
∴V/R=2*(V/Z)cosΦ
...(R remains same in both the cases)
∴Z/R=2cosΦ
Since R/Z is cos Φ, we can directly write,
2cos2Φ=1
∴cosΦ=1/√2=0.7071
(=√2/2 as given in the answer)

4. Feb 10, 2016

### Staff: Mentor

Nice. That works. It's great to see variations!

5. Aug 22, 2016

### fahraynk

Why do you set $I=\frac{V}{R-JX}$ instead of $\frac{V}{R+JX}$ In other words, why use the complex conjugate and not Z as it is?

6. Aug 22, 2016

### Staff: Mentor

Because the complex power is given by $P = V I^*$. See, for example: An Introduction to Electric Circuit Theory by G. Williams, section 5.8. Since the V that we are using here is purely real, the way to form the complex conjugate is to change the sign of the imaginary term in the impedance in the denominator.