Adjusting the Power Factor of an RLC circuit

  • #1
annamal
335
28
Homework Statement:
An RLC series circuit has an impedance of 60 ohms and a power factor of 0.50, with the voltage lagging the current. (a) Should a capacitor or an inductor be placed in series with the elements to raise the power factor of the circuit? (b) What is the value of the reactance across the inductor that will raise the power factor to unity?
Relevant Equations:
##P = 0.5\times IVcos(\phi)##

##Z = \sqrt{R^2 + (X_l - X_c)^2}##
a) An inductor should be added because that would cancel out the lag of the voltage with the current so that P = IV is at a maximum since V is ahead of I in an inductor
b) ##cos(\phi) = \frac{R}{Z}##, ##R = Zcos(\phi) = 60\times 0.5 = 30 ohms##
##X_l - X_c = \sqrt{Z^2 - R^2} = \sqrt{60^2 - 30^2}## = 52 ohms... don't know how to solve any longer
 

Answers and Replies

  • #2
alan123hk
726
409
Unity power factor means ##cos(\phi) ## needs to be equal to 1, so there is no lag or lead b/w voltage and current. 🤔
 
  • #3
DaveE
Science Advisor
Gold Member
2,841
2,488
Are you familiar with representing impedance with complex numbers, like ##Z_L = j \omega L## , or ##Z_C = \frac{1}{j \omega C}## ?
 
  • #4
annamal
335
28
Unity power factor means ##cos(\phi) ## needs to be equal to 1, so there is no lag or lead b/w voltage and current. 🤔
Yeah I get that. So R = Z, but I still don't know how to solve for ##X_L##
 
  • #5
annamal
335
28
Are you familiar with representing impedance with complex numbers, like ##Z_L = j \omega L## , or ##Z_C = \frac{1}{j \omega C}## ?
Yes, I am familiar with that, without the j that is.
 
  • #6
alan123hk
726
409
Yeah I get that. So R = Z, but I still don't know how to solve for ##X_L##

Now you know that you need to add a new inductor or replace the current one with a larger inductance, you also know that before adding the new inductor, the reactance difference between the original capacitor and the inductor is 52 ohms, I'm a little surprised, you still want to know What?

Is it the reactance of the newly added inductor or the total reactance of the newly added inductor in series with the original inductor? :rolleyes:
 
  • #7
berkeman
Mentor
64,182
15,410
Yes, I am familiar with that, without the j that is.
Wait, what? That makes no sense. And those equations make no sense without the "j" complex number qualifer.
 
  • #8
DaveE
Science Advisor
Gold Member
2,841
2,488
Yes, I am familiar with that, without the j that is.
So then, no. The ##j## is the whole point of complex numbers. ##j = \sqrt{-1}##, many people also use ##i = \sqrt{-1}## instead, it's the same thing. You WILL end up learning this at some point fairly soon, I think. It's the easy way to deal with circuit impedance.

That's OK, we just need to know to explain this the best way.

So, if the total (net) impedance seen by the source is inductive then the voltage leads the current. If it's capacitive the voltage lags the current. But for unity power factor the voltage should neither lead or lag the current, but be in phase. So given that the reactive impedance of a capacitor can cancel (or subtract from) the reactive impedance of an inductor (and vice-versa). What is the amount of additional reactance required to make the inductive reactance cancel the capacitive reactance?
 
  • #9
DaveE
Science Advisor
Gold Member
2,841
2,488
While you may not be familiar with complex numbers the reactance (impedance) of the capacitor works like this:
##Z_C = \frac{1}{j \omega C} = \frac{1}{\sqrt{-1} \omega C} = ( \frac{\sqrt{-1}}{\sqrt{-1}})( \frac{1}{\sqrt{-1} \omega C}) = - \sqrt{-1} \frac{1}{\omega C} = -j \frac{1}{ \omega C}##

Then you want ##|Z_C| = |Z_L| \Rightarrow \frac{1}{ \omega C} = \omega L## so that when you add them up ##Z_C + Z_L = -j \frac{1}{ \omega C} + j \omega L = j(- \frac{1}{ \omega C} + \omega L) = 0##

But then, of course, the total impedance of the series RLC is ##Z = R + Z_C + Z_L##
 
  • #10
annamal
335
28
Wait, what? That makes no sense. And those equations make no sense without the "j" complex number qualifer.
My book defines them here:
https://openstax.org/books/university-physics-volume-2/pages/15-2-simple-ac-circuits as equations 15.3 and 15.8
Screen Shot 2022-05-23 at 11.05.17 PM.png
Screen Shot 2022-05-23 at 11.05.53 PM.png
 
  • #11
annamal
335
28
While you may not be familiar with complex numbers the reactance (impedance) of the capacitor works like this:
##Z_C = \frac{1}{j \omega C} = \frac{1}{\sqrt{-1} \omega C} = ( \frac{\sqrt{-1}}{\sqrt{-1}})( \frac{1}{\sqrt{-1} \omega C}) = - \sqrt{-1} \frac{1}{\omega C} = -j \frac{1}{ \omega C}##

Then you want ##|Z_C| = |Z_L| \Rightarrow \frac{1}{ \omega C} = \omega L## so that when you add them up ##Z_C + Z_L = -j \frac{1}{ \omega C} + j \omega L = j(- \frac{1}{ \omega C} + \omega L) = 0##
Yes, and right now the ##Z_L - Z_C= 52##, so the new ##Z_L## has to be -52?
 
  • #12
DaveE
Science Advisor
Gold Member
2,841
2,488
Your book has buried the sign of the reactance into the words leading/lagging or something similar. They (or you) have left out the magnitude operator. What they should have said (or maybe did) was ## |\frac{V_0}{I_0}| = \frac{1}{\omega C} = X_C = |Z_C| ##.

PS: OK, plus, sorry to be snotty, but nobody should call a physics book "University Physics" and teach this subject without complex numbers. They are making it unnecessarily difficult by not waiting until you have the math background to make it easy to understand. This is not how this stuff is taught in the universities I am familiar with. JMHO.
 
Last edited:
  • #13
DaveE
Science Advisor
Gold Member
2,841
2,488
Yes, and right now the ##Z_L - Z_C= 52##, so the new ##Z_L## has to be -52?
Sorry, I haven't actually looked at the numbers. But...
You are being sloppy with your definitions. I don't understand "new ##Z_L##", give them different names or at least describe them with words. If it's a different inductor, it needs a different name. Do you mean you are adding an inductor with ##|Z_{L'}| = X_{L'} =-52##? How does that work? How do you get a negative value for ##|Z_{L'}| = \omega L' ##?
 
  • #14
annamal
335
28
Sorry, I haven't actually looked at the numbers. But...
You are being sloppy with your definitions. I don't understand "new ##Z_L##", give them different names or at least describe them with words. If it's a different inductor, it needs a different name. Do you mean you are adding an inductor with ##|Z_{L'}| = X_{L'} =-52##? How does that work? How do you get a negative value for ##|Z_{L'}| = \omega L' ##?
Yes, I am adding an inductor with ##|Z_{L'}| = X_{L'} = -52## so that ##X_L - X_C## = 0
 
  • #15
DaveE
Science Advisor
Gold Member
2,841
2,488
OK, I'll give you a break on the whole sign thing. Except to point out that magnitudes are never negative and inductance is never negative. Reactance can be negative, but that would be a capacitor with a non-negative capacitance. This is an important issue for you to get straight someday, but will be easier when you are comfortable with complex impedance values.

So, yes, adding a series inductor with a reactance X = +52, or and inductance of L = 52/ω will give you unity power factor.

But... Now that I've finally read the question carefully, I wonder if you copied it correctly? Part b) asks "What is the value of the reactance across the inductor that will raise the power factor to unity?". Compensating the reactance by adding a component "across the inductor" isn't a solvable problem without knowing the value of the original L or C, we only know the net reactance, the difference between the inductive and capacitive parts.

At best it's a poorly worded question. All we can know is that we have to add, not what the original individual reactive values were.
 
  • #17
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,697
1,274
Homework Statement:: An RLC series circuit has an impedance of 60 ohms and a power factor of 0.50, with the voltage lagging the current. (a) Should a capacitor or an inductor be placed in series with the elements to raise the power factor of the circuit? (b) What is the value of the reactance across the inductor that will raise the power factor to unity?
Relevant Equations:: ##P = 0.5\times IVcos(\phi)##

##Z = \sqrt{R^2 + (X_l - X_c)^2}##

a) An inductor should be added because that would cancel out the lag of the voltage with the current so that P = IV is at a maximum since V is ahead of I in an inductor
b) ##cos(\phi) = \frac{R}{Z}##, ##R = Zcos(\phi) = 60\times 0.5 = 30 ohms##
##X_l - X_c = \sqrt{Z^2 - R^2} = \sqrt{60^2 - 30^2}## = 52 ohms... don't know how to solve any longer
Be careful in solving this equation, ##~~\displaystyle Z = \sqrt{R^2 + (X_L - X_C)^2} ## ,

especially when solving for ##X_L-X_C## .

Squaring both sides gives : ##~~\displaystyle Z^2 = R^2 + (X_L - X_C)^2 ## .

Solving for ##(X_L-X_C)^2 ## gives: ##~~\displaystyle (X_L - X_C)^2 = Z^2 - R^2 ## .

When taking the square root of both sides, you must remember to supply a ± sign.

So you get ##~~\displaystyle X_L - X_C = \pm\sqrt{Z^2 - R^2~} ## .

The sign is determined by knowledge of what is lagging or leading.

Voltage versus Current
 
  • Like
Likes annamal and Delta2
  • #18
annamal
335
28
OK, I'll give you a break on the whole sign thing. Except to point out that magnitudes are never negative and inductance is never negative. Reactance can be negative, but that would be a capacitor with a non-negative capacitance. This is an important issue for you to get straight someday, but will be easier when you are comfortable with complex impedance values.

So, yes, adding a series inductor with a reactance X = +52, or and inductance of L = 52/ω will give you unity power factor.

But... Now that I've finally read the question carefully, I wonder if you copied it correctly? Part b) asks "What is the value of the reactance across the inductor that will raise the power factor to unity?". Compensating the reactance by adding a component "across the inductor" isn't a solvable problem without knowing the value of the original L or C, we only know the net reactance, the difference between the inductive and capacitive parts.

At best it's a poorly worded question. All we can know is that we have to add, not what the original individual reactive values were.
This is the problem exactly (#35):
Screen Shot 2022-05-24 at 11.47.02 PM.png
 
  • #19
annamal
335
28
Be careful in solving this equation, ##~~\displaystyle Z = \sqrt{R^2 + (X_L - X_C)^2} ## ,

especially when solving for ##X_L-X_C## .

Squaring both sides gives : ##~~\displaystyle Z^2 = R^2 + (X_L - X_C)^2 ## .

Solving for ##(X_L-X_C)^2 ## gives: ##~~\displaystyle (X_L - X_C)^2 = Z^2 - R^2 ## .

When taking the square root of both sides, you must remember to supply a ± sign.

So you get ##~~\displaystyle X_L - X_C = \pm\sqrt{Z^2 - R^2~} ## .

The sign is determined by knowledge of what is lagging or leading.

Voltage versus Current
I see, since voltage is lagging current, there is more of capacitor voltage and less of inductor voltage, leading to the negative sign of ##X_L - X_C##, so you have to add ##X_L## of +52 ohms
 
  • #20
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,697
1,274
I see, since voltage is lagging current, there is more of capacitor voltage and less of inductor voltage, leading to the negative sign of ##X_L - X_C##, so you have to add ##X_L## of +52 ohms
Yes.
 
  • #21
alan123hk
726
409
I'm also not sure if "the value of reactance across inductor L1" refers to the reactance of L1 itself or the reactance of another object externally connected to L1. But if it is the latter, there is no way to answer the question. So I have no choice but to assume that it actually refers to the reactance of L1 itself.

I see, since voltage is lagging current, there is more of capacitor voltage and less of inductor voltage, leading to the negative sign of XL−XC, so you have to add XL of +52 ohms
So I said in #6 that I wonder why you say you don't know how to solve ##X_L##, because you have calculated the correct answer in #1.
 

Suggested for: Adjusting the Power Factor of an RLC circuit

Replies
6
Views
300
Replies
4
Views
1K
  • Last Post
Replies
3
Views
543
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
10
Views
663
  • Last Post
Replies
2
Views
403
  • Last Post
Replies
4
Views
402
  • Last Post
Replies
2
Views
460
Top