Finding C & L from XC and XL

[Harley]

Hello everyone! I just joined this website after having a bad experience on a different forum.

First of all, I apologise if this thread is in the wrong place. I couldn't see a formula thread so figured general was the next logical step!

I am a student who is still learning the ropes, and was wondering if anyone could help me with the following.

I am trying to work out how I can calculate the capacitance and inductance from the capacitive reactance and inductive reactance. I know the formula for XL is 2 x pi x f x L and the formula for XC is 1/(2 x pi x f x c) but I am not yet confident in reversing formulas. I have tried a few different ways, but I wouldn't know if I'd got it right.

I have looked online, but the only relevant pages which appear are about calculating XL and XC not the other way around. I have also checked the books I have and can't find what I need.

I understand it is just switching the formulas around, but I want to make sure I do it correctly and if you don't ask questions you don't learn Could anyone please help me with this? How do I find C and L from XC and XL? I also have the frequency.

Thank you, the people on the other site were quite condescending and insulting, I'm hoping the people here are much kinder and helpful!

Thanks!

 

[davenn]

Hi ya Harley
welcome to EP

I am trying to work out how I can calculate the capacitance and inductance from the capacitive reactance and inductive reactance. I know the formula for XL is 2 x pi x f x L and the formula for XC is 1/(2 x pi x f x c) but I am not yet confident in reversing formulas. I have tried a few different ways, but I wouldn't know if I'd got it right.

yeah, my maths sux as well .... will see if some one else can help you out


Dave

 

[Harley]

Hi ya Harley
welcome to EP




yeah, my maths sux as well .... will see if some one else can help you out


Dave

Thanks a lot

 

[hevans1944]

Just use algebra. XL = 2 x pi x f x L requires you divide both sides of the equation by (2 x pi x f) to "cancel" those terms on the right side of the equation, leaving just L = XL / (2 x pi x f). Similarly for XC = 1 / (2 x pi x f x C) first take the reciprocal of both sides: 1 / XC = (2 x pi x f x C). Now divide both sides of the equation by (2 x pi x f) to leave C = 1 / ( XC x 2 x pi x f).

Oh, and welcome to EP, Harley.

 

[Harley]

Just use algebra. XL = 2 x pi x f x L requires you divide both sides of the equation by (2 x pi x f) to "cancel" those terms on the right side of the equation, leaving just L = XL / (2 x pi x f). Similarly for XC = 1 / (2 x pi x f x C) first take the reciprocal of both sides: 1 / XC = (2 x pi x f x C). Now divide both sides of the equation by (2 x pi x f) to leave C = (2 x pi x f) / XC.

Oh, and welcome to EP, Harley.

Thank you so much! I really appreciate your answer and kindness.

 

[hevans1944]

Thank you so much! I really appreciate your answer and kindness.

I had to correct the equation for C, My algebra must be rusty, Corrected result: C = 1 / (XC x 2 x pi x f). I have already edited my original post to show the correction. Sorry 'bout that.

 

[Harley]

I had to correct the equation for C, My algebra must be rusty, Corrected result: C = 1 / (XC x 2 x pi x f). I have already edited my original post to show the correction. Sorry 'bout that.

Not a problem! Thank you I was still a bit stuck but hopefully the edit fixes that!

To give a little more information, I've had to calculate the following things using the following formulas which was simple enough providing I've done it correctly!

Z = Square root of R(squared) + (XL - XC squared)
I = V/Z
VR = I x R
VL = I x XL
VC = I x XC

The part I am stuck on, is building the circuit and taking the measurements to prove my results are correct, only the thing is the software (Multisim) needs a capacitor and inductor value, hence my question.

 

[hevans1944]

I distrust circuit simulators and seldom use them, especially for simple RLC circuits. Best to learn how to use complex algebra and use that to check the simulator.

Note that, impedance is a complex quantity that has a real component and an imaginary component, expressed usually in the form of Z = R + jX where R is the real (power dissipating) resistive component and X is the reactive (non-power dissipating) component, and j = √ (-1) is the imaginary operator. By definition j² = -1. Both R and X are real numbers, but the complex sum Z = R + jX is not a real number. Unless R = 0 it is not a purely imaginary number either. The situation where X = 0 is called resonance, assuming both capacitive and inductive reactances exist. All of this complex number stuff affects how voltages and currents are calculated using the AC equivalent of Ohm's Law and Kirchoff's Laws.

All voltages and currents in an AC circuit that has inductance and/or capacitance, and also resistance, will be complex quantities. Complex arithmetic must be used to analyze such circuits. That can be a formidable task, so I can understand and even sympathize with the temptation to use a powerful simulator program to perform the grunt work.

In your example, I = V / Z is a complex quantity. Since Z varies with frequency, I will vary with frequency. That means VR, VL, and VC will all vary with frequency too. There are "special conditions" where |XL| = |XC| without regard to the fact the XL and XC are algebraic opposites. Such a condition is called resonance.

When you have an inductor in series with a capacitor, there is a frequency where the inductive reactance exactly cancels the capacitive reactance. The impedance of this series circuit is then purely resistive since the reactive or imaginary component is exactly zero. The resistance may be a specific part of the series circuit or it may be distributed, in the wire resistance of the inductor for example. Thus, you might connect a 100 Ω resistor in series with an "ideal" 1 H inductor and then connect the two in series with an "ideal" 1 μF capacitor. There will always be a frequency where the reactance of the inductor exactly cancels the reactance of the capacitor, leaving just the resistor to define the impedance at that resonant frequency. The reason this occurs is because inductive reactance is a monotonically increasing function of frequency, whereas capacitive reactance is a monotonically decreasing function of frequency. There will always be a frequency where these two functions are equal. That means that for any specific value of L and C there exists a resonant frequency.

You can use algebra to determine what the resonant frequency is when L and C are known. Just set XL = XC and solve the resulting equation for f:

2πfL = 1 / (2πfC) can be re-arranged to f² = 1 / (4π²LC) or f = 1 / (2π √LC). Plug in 1 H for L and 1 μF for C and you will find these two components will resonate at about 159 Hz. Note: the relation 2πf is called angular frequency, expressed in radians per second, and the symbol ω (lower-case omega) is used to represent it. So you can also state the ω = 1 / √(LC).

Are we having fun yet?

 

[hevans1944]

I just noticed that you use Z = √(R² + (XL - XC)²). This calculates the magnitude of Z but omits the angle that Z makes with the positive x-axis when impedance is plotted on Cartesian coordinates with the real component on the x-axis and the imaginary component on the y-axis. Just sayin'... it's something you will have to consider as you progress in your studies.

 

[davenn]

Just use algebra. XL = 2 x pi x f x L requires you divide both sides of the equation by (2 x pi x f) to "cancel" those terms on the right side of the equation, leaving just L = XL / (2 x pi x f). Similarly for XC = 1 / (2 x pi x f x C) first take the reciprocal of both sides: 1 / XC = (2 x pi x f x C). Now divide both sides of the equation by (2 x pi x f) to leave C = 1 / ( XC x 2 x pi x f).

Oh, and welcome to EP, Harley.

thanks Hop

I am absolutely useless at this sort of maths


Dave

 

[Harley]

Thank you, I followed your steps and got the exact same answers I got doing it the way I thought was wrong! I just have no confidence that is all it is, I was always bad at maths and can't get it into my head that I am improving.

The task has been solved with accurate results, thank you very much for your help!

In a similar LRC circuit, I am facing another problem, our assignments insist we verify our results using Multisim. The issue is, I have never had to measure the resonant frequency before and don't know how to go about it. I have made a separate post for it here: https://www.electronicspoint.com/th...sonant-frequency-fr-on-an-lrc-circuit.280972/ but I would certainly appreciate your help again if possible Thank you.

 

[Ratch]

Thank you, I followed your steps and got the exact same answers I got doing it the way I thought was wrong! I just have no confidence that is all it is, I was always bad at maths and can't get it into my head that I am improving.

The task has been solved with accurate results, thank you very much for your help!

In a similar LRC circuit, I am facing another problem, our assignments insist we verify our results using Multisim. The issue is, I have never had to measure the resonant frequency before and don't know how to go about it. I have made a separate post for it here: https://www.electronicspoint.com/th...sonant-frequency-fr-on-an-lrc-circuit.280972/ but I would certainly appreciate your help again if possible Thank you.

In a series resonant circuit, the frequency of resonance can be found by inserting a ammeter, and sweeping the circuit with a variable oscillator until the current peaks at maximum. The frequency that causes the current to peak is the resonant frequency. Otherwise, you can find the frequency where the voltage across the capacitor and coil each are equal. That will be the resonant frequency. Things are a lot different in a parallel resonant circuit.

Ratch

 

[hevans1944]

Well, Harley, (or Quinnzy as it may be), if you will go back to that other site you will find that somebody has posted a very detailed answer showing exactly how to do the necessary algebra. Perhaps you left that site too quickly.

I would have left a site that encouraged the posting of "condescending and insulting" responses too. And probably would never have gone back, even if someone here "recommended" it. For entirely different reasons, I no longer respond to questions on a popular site that solicits (and votes on) answers to a wide spectrum of questions, most of which are obvious homework assignments or trolls looking to start an argument. Not saying that doesn't occur here from time to time, but we don't do anyone's homework for them, and the moderators are really good at filtering out the trolls.

 

[hevans1944]

[quoted content removed ....]

She was given advice to the effect that if you intend to study electrical engineering, you need better (much better!) math skills. This is true, but hardly helpful to answering her question of how to solve for capacitance and inductance, given their reactances. And it is true that someone did eventually post the algebraic manipulations required. And the "answers" are voted and graded on that site, unlike here where all you can do is click on "Like" or move on to something else. On stackexchange you have to click-to-see-more to find the "answers" that drove Harley away. So, yeah, you need somewhat thick skin to deflect the "answers" that were not helpful. And I have a problem with stackexchange enforcing a "no dialog" policy. Their purpose is entirely different from the purpose of Electronics Point. We are here to encourage dialog and learning. Any "useful" answers are a sometimes happy by-product. Sorta reminds me of a large kindergarten play yard at times. With adult supervision of course.

 

[Harley]

This is exactly my point, I'm here to learn, not to be flamed for whatever reason, this thread has become off topic because yet again people can't just answer the question that's been asked. The posts I received from the people on there were unhelpful to say the least, I'm obviously going to go somewhere else seeking better help. Not to mention my question was "voted down"! What the Hell is that about? As no one answered, it's apparent people read my question, voted it down, and left without answering; if they are so clever, why didn't they answer it?

I requested my account be deleted within an hour of posting and looked for a different forum, there's nothing wrong with that; and yes, I did not see the answer there because I did not need to look for one there, I got an answer here right away.

Yes, I know my maths needs to improve, I am still a student, I am still learning, and I only learn what they teach unless I ask questions.

None of this is helpful nor constructive in regards to the topic of the thread.

Also, I'm not a he.

 

[Abcedfgh]

Hello everyone! I just joined this website after having a bad experience on a different forum.

First of all, I apologise if this thread is in the wrong place. I couldn't see a formula thread so figured general was the next logical step!

I am a student who is still learning the ropes, and was wondering if anyone could help me with the following.

I am trying to work out how I can calculate the capacitance and inductance from the capacitive reactance and inductive reactance. I know the formula for XL is 2 x pi x f x L and the formula for XC is 1/(2 x pi x f x c) but I am not yet confident in reversing formulas. I have tried a few different ways, but I wouldn't know if I'd got it right.

I have looked online, but the only relevant pages which appear are about calculating XL and XC not the other way around. I have also checked the books I have and can't find what I need.

I understand it is just switching the formulas around, but I want to make sure I do it correctly and if you don't ask questions you don't learn Could anyone please help me with this? How do I find C and L from XC and XL? I also have the frequency.

Thank you, the people on the other site were quite condescending and insulting, I'm hoping the people here are much kinder and helpful!

Thanks!

Okay so inductive reactance and capacitive reactance are opposites. So to find a point of unity you must first negate resistance. Use the xl=2 x pi x f x l to get inductive reactance

Now use the xc= 1/ 2 x pi x f x c

These two values are likely different but at one frequency will be equivalent.

Now let’s assume they aren’t. You can draw a quadrant grid with resistance being right of the origin. Conductance being left.

Inductive reactance is up and capacitive reactance is down.

Reactance and capacitance will cancel and equate to the subtraction of those numbers.
With vector sum addition (Pythagorean theorem) you can find the hypotenuse or z)

 

[Harald Kapp]

Old thread from 2016 -> closed