# RLC circuit confusion

1. Oct 5, 2012

### fawk3s

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

There are two circuits I dont entirely understand. Im also new to calculations with complex numbers and when to and when not to use them, so be easy on me. Anyways, the point is to find out what the different meters are showing, when you are given the initial voltage and the frequency of the current.

Here's the first one:
http://desmond.imageshack.us/Himg141/scaled.php?server=141&filename=21362017.png&res=landing [Broken]

So basically, its a circuit consisting of 2 resistors and an inductor. Whats interesting, but also confusing to me, is what the wattmeter is showing. The voltmeter U2, which is connected to basically the same point as the wattmeter, is showing 138 V. The ampermeter I1 is showing 1.24 A, which is measuring the same current as the wattmeter. That gives us the power P=1.24*138=171.12 W.
By doing some calculations I found out that 171.12 W is the power on the inductor and the resistor R2 together, on their impedance. But 92 W is the power on the resistor R2 alone.
So I guess my question here would be why is it so? How does the wattmeter work? I always thought that the wattmeter would measure the voltage and current, then somehow multiply them, and give us the result. But clearly, the multiplication of what the voltmeter and the ampermeter are showing us gives us a different result.
I do understand that the inductor preserves evergy as a magnetic field and then releases it, but Im still confused here. The voltmeter registers the voltage on the inductor, why doesnt the wattmeter?

The second situation I dont understand:
http://desmond.imageshack.us/Himg16/scaled.php?server=16&filename=ac2h.png&res=landing [Broken]

In this case, I dont really understand much. Im guessing you can add the capacities together, because they are in parallel. So that would give us a single 31 μF capacitor. But where to go from there? Im guessing we need the total impedance to find the current in I1. From there on Im guessing we could find U0 and U2. But how to find the impedance in this case? I tried a few ways but I guess they were wrong because the numers didnt match the ones in the picture. And again, why is the wattmeter showing the sum of the powers on resistors R2 and R3?

EDIT: for the ones who dont see the images for some reason:
1. http://desmond.imageshack.us/Himg141/scaled.php?server=141&filename=21362017.png&res=landing [Broken]
2. http://desmond.imageshack.us/Himg16/scaled.php?server=16&filename=ac2h.png&res=landing [Broken]

fawk3s

2. Relevant equations

Last edited by a moderator: May 6, 2017
2. Oct 5, 2012

### Staff: Mentor

The individual voltage and current meters are providing the amplitude (rms usually) of the voltage and current waveforms. However, because the circuits contain reactive components, the voltage and current waveforms are not in phase -- they reach their zeros and peaks at different times. Since the power is given by the multiplication of the voltage and current values at any given instant, and the peaks don't coincide, the amplitude of the power curve will not be the product of the peak values of each.

Wattmeters are clever beasts that will read the amplitude of the real power curve that results from the multiplication of the out of phase voltage and current waveforms.

3. Oct 5, 2012

### fawk3s

So, what you're saying is that volt and ampermeters measure the effective values, and so does the wattmeter, but since the wattmeter measures them at the same instant, it is able to detect the phase differences?
I've never really dived into exactly how the wattmeter works. Always figured it was just something that had amper and volt meter combined, and just "multiplied" the effective values, but apparently not. Since you seem to know about it, would you mind sharing some knowledge with me on what goes on inside the wattmeter?

(Yes, I do know what Wikipedia is, but most of the stuff on the net doesnt seem to have the necessary details Im looking for, so I'd really appreciate the help.)

4. Oct 5, 2012

### Staff: Mentor

The details depend upon the implementation of the wattmeter. There are various physical implementations to accomplish the result. Presumably the latest versions accomplish the measurements and multiplication digitally, then find the amplitude of the resulting curve.

Effectively, the wattmeter shows you the Real (or average) value of the power delivered, which is the power actually consumed by dissipative devices (resistors in this case).

On the other hand, multiplying the voltage and current meter readings yields the Apparent Power delivered, which is the magnitude of the complex power including the Real and Imaginary (reactive) components. The Real power is dissipated on resistances while the imaginary power is stored and returned by reactive components.

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5. Oct 6, 2012

### fawk3s

Im still not sure I understand. The average power dissipated in case of the waves being in phase is the multiplication of the effective values, right?

P = Imax/sqrt(2) * Umax/sqrt(2)

And this is exactly what the amper and volt meters are showing. So the only way that the wattmeter could be giving a different result is when it actually detects the phase differences. And that still raises the question on how it does it.
Or am I wrong somewhere or just ignorantly missing something?

And could somebody give some enlightment what goes on in the second circuit? I cant really see how Kirchhoff's first law applies there.