# Homework Help: AC circuit - voltage drop and resonance

1. Sep 18, 2014

### Rugile

1. The problem statement, all variables and given/known data

An AC voltmeter connected to the nodes A and B shows the same value as when connected to the nodes A and D (refer to attachment). What is the inductance of the inductor? What value does the voltmeter show? The inductor is ideal. U = 70sin(314t) V, C = 80 μF, R = 500 Ω.

2. Relevant equations

$U_{rms} = \frac{U_{max}}{\sqrt{2}}$
$Z_L = i \omega L; Z_C = -\frac{i}{\omega C}$
$\omega = \frac{1}{\sqrt{LC}}$

3. The attempt at a solution

For the voltages to be equal, the voltage drop across the inductor and capacitor in parallel should be 0. I know that in the case of series resonance, the voltage drop across the inductor and the capacitor is 0. Although here we also have a capacitor in parallel of the inductor, and due to the parallel resonance the voltage across these two elements is maximum (infinite). So what is confusing me, is that what is the actual voltage drop when we have simultaneous resonances? Could we say that the across the capacitor in series it is -∞ V, across the inductor ∞ V, thus resulting in 0 V in total (in series)?

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2. Sep 18, 2014

### ehild

The voltmeter reads the rms value of the voltage. If the readings are the same it means the rms values of the voltages are the same. The phases can be different.

In case L=0 the voltages UAB and UAD are the same. But that is not the only possibility.

ehild

Last edited: Sep 18, 2014
3. Sep 18, 2014

### Staff: Mentor

So what does this tell you about the voltage between B and D? How could you explain the occurrence of this value of voltage between B and D? What are its implications for the rest of the circuit?

4. Sep 18, 2014

### ehild

@NascentOxygen: Are you really suggesting that the voltage is zero between B and D????

ehild

5. Sep 18, 2014

### Staff: Mentor

Shhhhhh! Not so loud! It was a hint directed to Rugile, it's his homework task.

6. Sep 18, 2014

### Staff: Mentor

I think the intent of the problem would be to find the non-trivial solution, or perhaps both the trivial and non-trivial solutions

7. Sep 18, 2014

### ehild

What you suggested was the first idea of the OP :

which is wrong. If UBD is zero, UAB=UAD, but from the fact that the voltmeter measures equal voltages between A , B and A , D does not follow that UBD=0. Can you imagine something else, not so trivial?

ehild

8. Sep 18, 2014

### Rugile

Thank you for all the answers!

So it does in fact have nothing to do with the resonance?

Why is the voltage the same when L = 0? Is it because of the phase shift?

This is really confusing! I don't really see any other way than UBD being zero. It seems pretty reasonable to me!

9. Sep 18, 2014

### ehild

It is not connected to resonance.

If L=0 the impedance of the inductor is zero. Connected parallel with C, the resultant is also zero. You can replace the parallel connected inductor and capacitor by a single wire of zero resistance connecting the points B and D. . There is no voltage drop across it, so the potential is the same at both points. This is the trivial solution of the problem.

The impedances are complex quantities and so are the voltages.

The complex impedance of the parallel LC is ZLC and that of the series R, C is ZRC, $U_{AD}=U_{AB}\frac{Z_{RC}}{Z_{LC}+Z_{RC}}$

UAD is not in phase with UAB, but the voltmeter reads only the rms values. The two readings are equal - it means that the rms value (or the amplitude) of UAD is equal to those of UAB.
If $|U_{AD}|=|U_{AB}|$

$|\frac{Z_{RC}}{Z_{LC}+Z_{RC}}|=1$ follows.

Find L which satisfies the condition above.

ehild

10. Sep 18, 2014

### Rugile

..... true.

Thank you a lot!!

11. Sep 18, 2014

### ehild

Could you figure out L?

ehild

12. Sep 18, 2014

### Rugile

Ok, so we get $Z_{RC} = Z_{RC} + Z_{LC}$, right?
Then $Z_{LC} = 0 = \frac{-i \omega L * \frac{i}{\omega C}}{i \omega L - \frac{i}{\omega C}} = \frac{i \omega L}{1 - \omega^2 LC}$ and, well, the answer - L=0...

13. Sep 18, 2014

### ehild

That is the trivial case. But the voltage readings are the same if the magnitude of ZRC is equal to the magnitude of ZRC+ZLC, that is

$\big|R-\frac{j}{ωC} \big|=\big|R-\frac{j}{ωC}+\frac{jωL}{1-ω^2LC}\big|$

ehild

14. Sep 19, 2014

### Rugile

Okay, so Zlc = -2Zrc ( if we say that Zlc is negative and the modulus of difference of Zrc and Zlc has to be equal to Zrc, Zlc has to be twice as big). Then after rearranging we get: $L=\frac{2(i-R\omega C)}{3i \omega^2 C}$. What do we do with the i? Should it be rearranged to L=a + ib, and L =√(a^2 + b^2)?

15. Sep 19, 2014

### Staff: Mentor

You included the word "resonance" in the thread's subject---is this question from the the resonance topic of your textbook?

Does the textbook give you the answer to this question?

http://imageshack.com/a/img440/6184/icon2p.gif [Broken] Could you check back with the original question, and confirm that it states voltages AB and AD measure as the same value.

* If the question is as you say, then my earlier hints can be ignored, because I had inadvertently swapped A and B. Such a swap is logical for a question dealing with resonance, IMO.

Last edited by a moderator: May 6, 2017
16. Sep 19, 2014

### Rugile

The problem is not from a textbook - I don't have the answer (I didn't make it up myself, either ) - and it's under a more broad topic - electromagnetism. I just though that resonance is probably the key to the problem.

And yes, the problem statement is correct in the first post.

17. Sep 19, 2014

### ehild

NO, L is real.

The impedances are complex quantities, Z=a+ib, and you get the magnitude (modulus) or absolute value of the impedance as $|Z|=\sqrt{a^2+b^2}$

ZLC is imaginary, ZRC is complex.

Collect the real and imaginary terms on the right side and take the magnitude in both sides of the equation

$\big|R-\frac{j}{ωC} \big|=\big|R-\frac{j}{ωC}+\frac{jωL}{1-ω^2LC}\big|$

ehild

Calculate what would an AC voltmeter measure between A and D if ωL=2/3 (1/(ωC))

Last edited: Sep 19, 2014
18. Sep 19, 2014

### Staff: Mentor

I haven't looked at the maths, but from the circuit I anticipate there to be 3 correct answers, one of which is L=0.

19. Sep 20, 2014

### Rugile

Got it...

$|R- j\frac{1}{\omega C}| = | R - j(\frac{-1}{\omega C} + \frac{\omega L}{1 - \omega^2 L C} | => \sqrt{R^2 + (\frac{1}{\omega C})^2 } = \sqrt{ R^2 + ( \frac{\omega L}{1 - \omega^2 LC} - \frac{1}{\omega C} )^2 }$


20. Sep 20, 2014

### Rugile

Accidentally posted the post above before finishing it, and for some reason can't edit it, so I'll finish here:

$|R- j\frac{1}{\omega C}| = | R - j(\frac{-1}{\omega C} + \frac{\omega L}{1 - \omega^2 L C} | => \sqrt{R^2 + (\frac{1}{\omega C})^2 } = \sqrt{ R^2 + ( \frac{\omega L}{1 - \omega^2 LC} - \frac{1}{\omega C} )^2 }$
$\frac{1}{\omega^2 C^2} = (\frac{\omega L}{1-\omega^2 LC} - \frac{1}{\omega C})^2$
$\frac{1}{\omega C} = \frac{\omega L}{1-\omega^2 LC} - \frac{1}{\omega C}$
$L = \frac{2}{3 \omega^2 C}$

Thank you for the help (and patience)!!

So is there one more answer?

21. Sep 20, 2014

### ehild

$$\frac{1}{\omega^2 C^2} =\left( \frac{\omega L}{1-\omega^2 LC} - \frac{1}{\omega C}\right)^2.$$ has two roots: L=0 and L=2/3 1/(w^2C)

ehild