# RLC circuit

## Homework Statement

See the picture. If at electrical resonance ##U_c=60 V## and ##U_L=40 V##, find the ratio of reactances ##\frac{X_L}{X_c}##.

## Homework Equations

[/B]##Z=\sqrt{R^2+(\omega L - \frac{1}{\omega C})^2}## , ## Y= \sqrt{\frac{1}{R^2} + (\omega C - \frac{1}{\omega L})^2}## , ##\phi = \arctan \frac{\omega L - \frac{1}{\omega C}}{R}## and so on

## The Attempt at a Solution

[/B]
I am not sure how exactly use given conditions, I somehow got that if there is electrical resonance , than
##R^2C=R^2C^2\omega^2L+L## but that's surely wrong, and not sure how to use the voltages given.

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BvU
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"I somehow got" is not very strong. What does your textbook or your notes say ?

ehild
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## Homework Statement

See the picture. If at electrical resonance ##U_c=60 V## and ##U_L=40 V##, find the ratio of reactances ##\frac{X_L}{X_c}##.

## Homework Equations

[/B]##Z=\sqrt{R^2+(\omega L - \frac{1}{\omega C})^2}## , ## Y= \sqrt{\frac{1}{R^2} + (\omega C - \frac{1}{\omega L})^2}## , ##\phi = \arctan \frac{\omega L - \frac{1}{\omega C}}{R}## and so on
These equations are not relevant for the circuit in the problem.

## The Attempt at a Solution

[/B]
I am not sure how exactly use given conditions, I somehow got that if there is electrical resonance , than
##R^2C=R^2C^2\omega^2L+L## but that's surely wrong, and not sure how to use the voltages given.
What is the current at resonance? What are the expressions for UL and Uc at resonance?

• BvU
These equations are not relevant for the circuit in the problem.
Maybe not these in particular, but little modified,I used them to get ## R^2C=R^2C^2\omega^2 L + L ##. I got that in two different ways (then I thought that it might be correct).
I first calculate ## Z_{RC}## (parallel) with ##Y_{RC}## and since I know ##\tan \phi##, i can calculate ## \sin \phi=\frac{\tan \phi}{\sqrt{\tan^2 \phi+1}}##, and since imaginary part of ##Z_{RLC}## has to be ##0##, then ##Z_{RC} \sin \phi = L \omega## . The other way was to calculate ##Z_{RC}= \frac{-jRC \omega}{R-jC\omega}## and so on. Is there maybe easier way to derive this formula?

What is the current at resonance? What are the expressions for UL and Uc at resonance?
I am not sure, this isn't only serial or parallel circuit. Maybe
## U_L=I_{ef} \cdot X_{L} ##
## U_C=I_{ef} \cdot \frac{R}{X_C+R} \cdot X_{C}## ?

ehild
Homework Helper
Maybe not these in particular, but little modified,I used them to get ## R^2C=R^2C^2\omega^2 L + L ##. I got that in two different ways (then I thought that it might be correct).
I first calculate ## Z_{RC}## (parallel) with ##Y_{RC}## and since I know ##\tan \phi##, i can calculate ## \sin \phi=\frac{\tan \phi}{\sqrt{\tan^2 \phi+1}}##, and since imaginary part of ##Z_{RLC}## has to be ##0##, then ##Z_{RC} \sin \phi = L \omega## . The other way was to calculate ##Z_{RC}= \frac{-jRC \omega}{R-jC\omega}## and so on. Is there maybe easier way to derive this formula?
I do not think there is an easier way. I calculated the imaginary part of ZRC and added to ωL and made the sum equal to zero. But your formula for ZRC is not correct. Remember that Xc=-j/(ωC).
Knowing the resonance frequency, you can determine the impedance at resonance, and the current in terms of the generator voltage.

I am not sure, this isn't only serial or parallel circuit. Maybe
## U_L=I_{ef} \cdot X_{L} ##
## U_C=I_{ef} \cdot \frac{R}{X_C+R} \cdot X_{C}## ?
It is all right. Use the result for the resonant frequency and current.

• crom1
Yes, you are right, I meant ##Z_{RC}=\frac{-jRX_{C}}{R-jX_{C}}##.
I got that ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}## and from voltages that

## \frac{U_{L}}{U_{C}}=\frac{2}{3} = \frac{X_L(X_C+R)}{RX_C} ##. Don't know what to do next.

ehild
Homework Helper
Yes, you are right, I meant ##Z_{RC}=\frac{-jRX_{C}}{R-jX_{C}}##.
I got that ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}## and from voltages that

## \frac{U_{L}}{U_{C}}=\frac{2}{3} = \frac{X_L(X_C+R)}{RX_C} ##. Don't know what to do next.
You need to eliminate R somehow. Use the expression of the resonant frequency.
At resonance, Z is real. Write up the complex impedance and take the real part. It becomes quite simple.

• crom1
Well,I did got ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}+j\frac{R^2X_L-R^2X_C+X_LX_C^2}{R^2+X_{C}^2} \Rightarrow Z=\frac{RX_{C}^2}{R^2+X_{C}^2} ##
Now, is omega ##\omega= \frac{1}{RC} \sqrt{\frac{R^2C}{L}-1} ##?
When you say resonant frequency do you mean ## f=\frac{\omega}{2\pi}## ?
And if yes, how should I use that expression?

ehild
Homework Helper
Well,I did got ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}+j\frac{R^2X_L-R^2X_C+X_LX_C^2}{R^2+X_{C}^2} \Rightarrow Z=\frac{RX_{C}^2}{R^2+X_{C}^2} ##
Correct. Substitute the expression for Xc.
Now, is omega ##\omega= \frac{1}{RC} \sqrt{\frac{R^2C}{L}-1} ##?
When you say resonant frequency do you mean ## f=\frac{\omega}{2\pi}## ?
And if yes, how should I use that expression?
I mean the angular frequency, ω, when I say frequency. Use the expression for ω to get the impedance at resonance.

• crom1
I am probably missing something obvious here,but how do I get impedance with ##\omega##?

From those voltage equations I got that ##R=\frac{3X_LX_C}{2X_C-3X_L}## and I tried to use that in ##R^2X_L-R^2X_C+X_LX_C^2=0## but didn't lead me nowhere.

Last edited:
ehild
Homework Helper
How is XC and XL defined?

Is ##\omega= \frac{1}{RC} \sqrt{\frac{R^2C}{L}-1} ##?
I have impedance ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}## and (if thats useful at all) ##R=\frac{3X_LX_C}{2X_C-3X_L}##.
##X_L = L \omega## and ## X_C = \frac{1}{C \omega}##.
I don't know how to find impedance with ##\omega##.

ehild
Homework Helper
Is ##\omega= \frac{1}{RC} \sqrt{\frac{R^2C}{L}-1} ##?
yes.
I have impedance ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}## and (if thats useful at all) ##R=\frac{3X_LX_C}{2X_C-3X_L}##.
##X_L = L \omega## and ## X_C = \frac{1}{C \omega}##.
I don't know how to find impedance with ##\omega##.
XC=-1/(ωC). Sustitute the expressions for XL and XC into the equation for Z. Then substitute the expression of ω.

Okay I got ##Z=\frac{L}{RC}## . How do I use that ?
Since ## \frac{X_L}{X_C}= LC\omega ^2 = 1-\frac{L}{R^2C} = 1-\frac{Z}{R}## , does it have to do something with this?

ehild
Homework Helper
Okay I got ##Z=\frac{L}{RC}## . How do I use that ?
Since ## \frac{X_L}{X_C}= LC\omega ^2 = 1-\frac{L}{R^2C} = 1-\frac{Z}{R}## , does it have to do something with this?
You know Z . What is the generator current? With the current, you get an equation for UL and UC.

You mean
##U_L=I_{ef} \cdot X_{L} ##
##U_C=I_{ef} \cdot \frac{R}{X_C+R} \cdot X_{C} ##

where ## I_{ef}= \frac{U_{ef}}{Z} ##?

ehild
Homework Helper
You mean
##U_L=I_{ef} \cdot X_{L} ##
##U_C=I_{ef} \cdot \frac{R}{X_C+R} \cdot X_{C} ##

where ## I_{ef}= \frac{U_{ef}}{Z} ##?
You know that the effective voltages are UL=IeffωL=40 V and UC=60 V.
But you get UC if you multiply Ieff with the magnitude of the impedance ZRC which is not RXc/(R+Xc).
Write the equations for UC and UL in terms of C, L, R and Ueff.
You have to get a numercal value for XC/XL.

• crom1
ehild
Homework Helper
You can not get the complex impedance of the parallel resistor and capacitor correctly. The impedance of the capacitor is ##\frac{1}{j\omega C} ##, so ##Z_{RC}=\frac{1}{jωC+\frac{1}{R}}=\frac{R}{1+jωCR}##.
Its magnitude is ##|Z_{RC}|=\frac{R}{\sqrt{1+(ωCR)^2}}##
In #Post 8, you got that the impedance is
##
Z=\frac{RX_{C}^2}{R^2+X_{C}^2}
## at resonance. Substitute ##X_C=-\frac{1}{ωC}##, you get ##
Z=\frac{R}{(RωC)^2+1}
##
Compare it with |ZRC|.
You also derived that the impedance at resonance is ##Z=\frac{L}{RC}##, so you can write |ZRC| in terms of L and C. Use the equations for UL and UC and divide them. What do you get?

• crom1
Hi, sorry I wasn't active, I decided to practice these kind of problems more (easier problems) and then try to solve this one. I got by drawing vector diagram that:

and ##U_C## vector has angle ##\sin \phi= \frac{-U_L}{U_C} \rightarrow \sin \phi = -\frac{2}{3}##.
And ## \sin | \phi| = \frac{I_C}{I}## so ##\frac{I }{I_C} = \frac{3}{2} ##

Then ## \frac{X_C}{X_L}= \frac{I}{I_C} \cdot \frac{U_C}{U_L} = \frac{9}{4}=2.25##

Is this correct?

ehild
Homework Helper
It is correct, and a very nice short solution. Congratulation! I prefer using complex impedances instead of vector diagrams, but my solution was very long and complicated.

ehild
Homework Helper
Hi, sorry I wasn't active, I decided to practice these kind of problems more (easier problems) and then try to solve this one. I got by drawing vector diagram that:

and ##U_C## vector has angle ##\sin \phi= \frac{-U_L}{U_C} \rightarrow \sin \phi = -\frac{2}{3}##.
And ## \sin | \phi| = \frac{I_C}{I}## so ##\frac{I }{I_C} = \frac{3}{2} ##

Then ## \frac{X_C}{X_L}= \frac{I}{I_C} \cdot \frac{U_C}{U_L} = \frac{9}{4}=2.25##

Is this correct?
To make the solution more clear, I add the phasor diagram. Because of resonance, the generator voltage U is in phas with the generator current I. 