Adjusting the Power Factor of an RLC circuit

In summary,If the power in a circuit is being divided between two components, one of which is an inductor, the voltage will always be ahead of the current in the inductor, meaning that the power P = IV is at its maximum. This is why an inductor is needed to cancel out the lag of the voltage with the current so that P = IV is at a maximum.
  • #1
358
31
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
 
Physics news on Phys.org
  • #2
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
Are you familiar with representing impedance with complex numbers, like ##Z_L = j \omega L## , or ##Z_C = \frac{1}{j \omega C}## ?
 
  • #4
alan123hk said:
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
DaveE said:
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
annamal said:
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
annamal said:
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
annamal said:
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
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
berkeman said:
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
DaveE said:
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
annamal said:
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:
  • Like
Likes berkeman
  • #13
annamal said:
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
DaveE said:
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
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.
 
  • Like
Likes Delta2
  • #17
annamal said:
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
DaveE said:
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
SammyS said:
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
 
  • Like
Likes DaveE
  • #20
annamal said:
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.
 
  • Like
Likes alan123hk
  • #21
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.

annamal said:
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
5
Views
220
Replies
12
Views
180
Replies
10
Views
436
Replies
2
Views
170
Replies
3
Views
1K
Replies
12
Views
3K
Replies
4
Views
2K
Replies
2
Views
744
Replies
2
Views
788
Back
Top