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Resonance in a circuit with L and C

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  • Thread starter Korisnik
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  • #1
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1

Homework Statement



http://i.imgur.com/FxuxHjW.png?1
L = 1.6 mH, C = 100 μF. Determine the frequency at which the total current will be 0. (The picture shows 3L, which is 3*L, so 3*1.6 mH.) There's also a hint "Use the vector diagram." The result is 2.5 kHz which I cannot get.

2. Homework Equations

[tex]f=\frac{1}{2\pi\sqrt{L\cdot C}}[/tex]

The Attempt at a Solution



I tried simply plugging the numbers in the formula but it doesn't work. :) I think I don't understand what's going on to actually solve the problem. I think what happens in one parallel is: current will be 0 because the values of capacitance and inductance are of the same size but of opposite signs, so there are going to be two currents which cancel out; some books say there's also infinite resistance (by the formula R_parallel = R_1*R_2/(R_1 + R_2), so I'm guessing, since X_C = - X_L, when you add them, the denominator is 0 which implies infinite resistance)...

I've never seen such a problem... Also, I'm not sure how to solve this using a phasor diagram as I cannot actually represent L, C, or reactances themselves... (only voltages and currents).

Thanks in advance.
 

Answers and Replies

  • #2
gneill
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Hi Korisnik. Your Relevant equation applies to case of a circuit with a single L and a single C, so either a simple LC series or LC parallel circuit. Here the situation is a bit more complicated.

One approach you might consider is to use impedance (complex valued "resistance") and admittance (complex valued "conductance") to find an expression for the overall impedance or admittance of the load with respect to angular frequency ##\omega##. Have you been introduced to impedance and admittance yet?
 
  • #3
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Hi Korisnik. Your Relevant equation applies to case of a circuit with a single L and a single C, so either a simple LC series or LC parallel circuit. Here the situation is a bit more complicated.

One approach you might consider is to use impedance (complex valued "resistance") and admittance (complex valued "conductance") to find an expression for the overall impedance or admittance of the load with respect to angular frequency ##\omega##. Have you been introduced to impedance and admittance yet?
Yes, I know about these, but I'm afraid I don't quite know how to apply that here. I'd use impedance since admittance is simply useless (extra symbols for 1/x thing). If I sum all of the reactances: -jX_C || jX_3L + -jX_C || jX_3L + jX_3L = 0 I get: 3*3L/C +omega^2 * 9L^2 = 0; plugging in numbers and I got the freakin' solution. I don't know why I never listen to myself... I did this halfway and said it's too easy and never completed it. Thanks. :)

EDIT: although I got omega = 2500 r/s, so i believe the answer provided is wrong... (the result divided by 2pi should do it).
 
  • #4
BvU
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Not so fast ! In the picture, the left 3L is not in series with the two L//C circuits !
 
  • #5
gneill
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I'd use impedance since admittance is simply useless
Not so! When the admittance of the load is zero then no current can flow, which is the condition you're looking for.
 
  • #6
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Not so fast ! In the picture, the left 3L is not in series with the two L//C circuits !
F***, ok I get then B_L = B_C, which is the same as X_L=X_C, and the result is wrong.
 
  • #7
BvU
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F*** , ok I get then B_L = B_C, which is the same as X_L=X_C, and the result is wrong.
I'm not familiar with your notation. Also the vector picture thing is somewhat outlandish to me. I am used to working with impedances. They work just like resistances, but have an imaginary part as well. Is that ok with you ? Since you mention j in post #3 I suppose so.

For one L and one C in parallel we have ##Z^{-1} = \displaystyle {1\over j\omega L} + j\omega C\;##. Put two of these in series, you get two times that ##Z##. Call it ##Z_s##.

For the left L in the picture and ##Z_s## in parallel, we have ##{Z_{total}}^{-1} = \displaystyle {1\over j\omega L} + Z_s^{-1}##. You want ##{Z_{total}}^{-1} = 0## to get i = 0.

My answer then happens to be the same as what you state in your original posting (with L = 1.6 mH). However, it comes out a factor 9 lower than the given result.
 
  • #8
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I'm not familiar with your notation. Also the vector picture thing is somewhat outlandish to me. I am used to working with impedances. They work just like resistances, but have an imaginary part as well. Is that ok with you ? Since you mention j in post #3 I suppose so.

For one L and one C in parallel we have ##Z^{-1} = \displaystyle {1\over j\omega L} + j\omega C\;##. Put two of these in series, you get two times that ##Z##. Call it ##Z_s##.

For the left L in the picture and ##Z_s## in parallel, we have ##{Z_{total}}^{-1} = \displaystyle {1\over j\omega L} + Z_s^{-1}##. You want ##{Z_{total}}^{-1} = 0## to get i = 0.
Now I'm not sure if I'm doing it right; impedance is admittance ^-1, but shouldn't admittance in series be exactly what normal impedance is in parallel? So when you have 2 admittances in series they should add like R1*R2/(R1+R2)... So I think that's where you mistook. Having said that, lemme try again and calculate it. Also, B is notation for susceptance:
First I'll change names B_L = L and B_C = C, || means parallel and (+) series:
I have [tex]-jL || (-jL||jC (+) -jL||jC) = 0
\\-jL||(\frac{(jC-jL)^2}{2(jC-jL)}) = 0\\ -jL+\frac{jC-jL}{2} =0\\ 3jL=jC[/tex]
Now when you substitute back: L = B_L = 1/(omega*L), and C=B_C = C*omega, you get the same formula but this time with a sqrt(3) in the numerator, and the result is: omega = 2500, which is true, but it's not a frequency, so it should be divided by 2pi...
 
  • #9
gneill
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If you're uncomfortable summing admittances in series you can always sum impedances first, then take the reciprocal. That is, Y = 1/Z.

Omega is certainly a frequency. It just happens to be an angular frequency. In many ways it is more fundamental than the Hertzian frequency f = 1/period. It may well be that the author of the problem meant for you to find the angular frequency but failed to specify it.
 
  • #10
rude man
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My answer is about 1/5 of 2.5 KHz. But, lots of opportunity for mistakes.
I'm not familiar with your notation. Also the vector picture thing is somewhat outlandish to me. I am used to working with impedances. They work just like resistances, but have an imaginary part as well. Is that ok with you ? Since you mention j in post #3 I suppose so.

For one L and one C in parallel we have ##Z^{-1} = \displaystyle {1\over j\omega L} + j\omega C\;##. Put two of these in series, you get two times that ##Z##. Call it ##Z_s##.

For the left L in the picture and ##Z_s## in parallel, we have ##{Z_{total}}^{-1} = \displaystyle {1\over j\omega L} + Z_s^{-1}##. You want ##{Z_{total}}^{-1} = 0## to get i = 0.
The left L in the picture is 3L.
 
  • #11
rude man
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I got f ~ 1/5 of 2.5 KHz. But, lots of possibilities for mistakes.
 
  • #12
BvU
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...
I have [tex]-jL || (-jL||jC (+) -jL||jC) = 0
\\-jL||(\frac{(jC-jL)^2}{2(jC-jL)}) = 0\\ -jL+\frac{jC-jL}{2} =0\\ 3jL=jC[/tex]
Now when you substitute back: L = B_L = 1/(omega*L), and C=B_C = C*omega, you get the same formula but this time with a sqrt(3) in the numerator, and the result is: omega = 2500, which is true, but it's not a frequency, so it should be divided by 2pi...
I agree with the calculation and the result. At first I found the notation utterly confusing, but with the ##L = 1/(\omega L)## substitution it comes out alright. However, you run the risk of confusing yourself too with this kind of notation. And on top of all that then follows an equally confusing L = 3L !
I would be in favour (in all modesty) of a notation like $${1\over Z} = {1\over Z_L}+{1\over Z_C} = {1\over j\omega L}+ j\omega C$$Two of these ##Z## in series:$${1\over Z_S} = {1\over 2}\left ( {1\over Z_L}+{1\over Z_C}\right ) = {1\over 2 j\omega L}+ {j\omega C\over 2}$$That lot in parallel with the left induction:$${1\over Z_{Total}} = {1\over Z_L} + {1\over 2 j\omega L}+ {j\omega C\over 2}= {3\over 2 j\omega L}+ {j\omega C\over 2}={j\over 2}\left (\omega C - {3\over \omega L}\right )$$ So that $$ {1\over Z_{Total}} = 0 {\rm\ \ if\ \ } f={1\over 2\pi\sqrt{LC\over 3}}$$And with L/3 = 1.6 mH and C 100 μF I get 398 Hz (2500 rad/s / (2 π) ) too.
 
  • #13
rude man
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And with L/3 = 1.6 mH and C 100 μF I get 398 Hz (2500 rad/s / (2 π) ) too.
But, L = 1.6 mH, not L/3 = 1.6 mH.
Keep at it until you get 513.9 Hz. (How's that for temerity! :smile: )
 
  • #14
BvU
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Hehe, you see it's confusing ! For each of the three inductances in the diagram the inductance is 4.8 mH. As you pointed out yesterday. I don't want to type ## {1\over Z_L} = {1\over j\omega 3L} ## in order not to make it worse for others.
 
  • #15
rude man
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Yeh, but we ought to be able to agree on the final answer ...
 
  • #16
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PF shies away from final answers, but since Kori worked it out in full, I didn't mind confirming it with a calculation that I hope is less confusing. Same result: ##\omega## = 2500 rad/s. I think the 513.9 Hz still has a little mistake in it, but I'm open for convincing arguments :)
 
  • #17
rude man
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PF shies away from final answers, but since Kori worked it out in full, I didn't mind confirming it with a calculation that I hope is less confusing. Same result: ##\omega## = 2500 rad/s. I think the 513.9 Hz still has a little mistake in it, but I'm open for convincing arguments :)
I see a mistake I made. I'll recompute & report back, probably to confirm your number.
 
  • #18
rude man
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I confirm f = 398 Hz. Whew!
 
  • #19
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I think we've lost Kori. But at least we all agree on the numerical result. Now I would like someone to enlighten us with a response that makes use of the vector diaggram the that was hinted at in the OP. Anyone ?
 
  • #20
rude man
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I think we've lost Kori. But at least we all agree on the numerical result. Now I would like someone to enlighten us with a response that makes use of the vector diaggram the that was hinted at in the OP. Anyone ?
I pass! :s
 

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