• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Archived Power factor at half power points for RLC circuit

  • Thread starter Loki
  • Start date

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:

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

This seems correct to me, but the book says the answer is [itex]\frac{\sqrt {2}}{2}[/itex]

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:

gneill

Mentor
20,489
2,615
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}##.
 

cnh1995

Homework Helper
Gold Member
3,280
1,054
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)
 

gneill

Mentor
20,489
2,615
Nice. That works. It's great to see variations!
 
184
5
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}##.
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?
 

gneill

Mentor
20,489
2,615
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?
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.
 

Want to reply to this thread?

"Power factor at half power points for RLC circuit" You must log in or register to reply here.

Related Threads for: Power factor at half power points for RLC circuit

Replies
1
Views
29K
  • Posted
Replies
8
Views
14K
  • Posted
Replies
1
Views
988
  • Posted
Replies
3
Views
13K
  • Posted
Replies
1
Views
3K
  • Posted
Replies
12
Views
19K
Replies
2
Views
5K
  • Posted
Replies
11
Views
12K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top