Finding Reactance Ratio in Electrical Resonance

[crom1]

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.

 

[BvU]

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

 

[ehild]

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.

Your equation is correct.
What is the current at resonance? What are the expressions for U_L and Uc at resonance?

 

[crom1]

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?

Your equation is correct.
What is the current at resonance? What are the expressions for U_L 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]

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 Z_RC and added to ωL and made the sum equal to zero. But your formula for Z_RC 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]

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]

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.

 

[ehild]

How is X_C and X_L defined?

 

[crom1]

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 that's 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]

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 that's 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 $\omega$.

X_C=-1/(ωC). Sustitute the expressions for X_L and X_C into the equation for Z. Then substitute the expression of ω.

 

[crom1]

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]

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 U_L and U_C.

 

[crom1]

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]

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=I_effωL=40 V and U_C=60 V.
But you get U_C if you multiply I_eff with the magnitude of the impedance Z_RC which is not RXc/(R+Xc).
Write the equations for U_C and U_L in terms of C, L, R and Ueff.
You have to get a numercal value for X_C/X_L.

 

[ehild]

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 |Z_RC|.
You also derived that the impedance at resonance is $Z=\frac{L}{RC}$, so you can write |Z_RC| in terms of L and C. Use the equations for U_L and U_C 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]

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]

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.