# Series/Parallel RLC

1. Jun 22, 2009

### Petrucciowns

I'm typing up all of my old homework assignments into notes and have forgotten quite a bit of things since I have learned them.

My problem now is with a series/parallel RLC circuit. I have no problem with series or parallel, but this combined one is giving me difficulties.

I am able to solve several of the questions once ZT is already found such as:
IT= source/ZT VR1= ITxR

,but the rest of the circuit has my mind boggled. I know with series/parallel resistive circuits you can subtract the source from the series resistor to get the voltage across the parallel section. It seems that this is not the case here.

So what I am having difficulties with are: ZT, VR2, IL, IC, PT, P.F.

I would appreciate any help. Click the picture to zoom.

http://img193.imageshack.us/img193/994/36987056.jpg [Broken]

Last edited by a moderator: May 4, 2017
2. Jun 22, 2009

### cepheid

Staff Emeritus
Once you convert everything into impedances, you can analyze the circuit just as you would with a resistor network (bearing in mind that all of your measured quantities are phasors). So, let me get you started:

ZT = ZR1 + [ ZL || (ZR2 + ZC) ] ​

Now what is the impedance of a capacitor? How about an inductor?

3. Jun 22, 2009

### Petrucciowns

Well the impedence of the capacitor and inductor are already shown. What I would do is find the parallel section by taking:

1/20 to the angle of 90 + 1/30 to the angle of -90 Then 1/ the sum

Then the sum + R2+ R1.

But I never come up with the right answer for ZT. I know the attached image file has the correct answer for ZT ,because we graded it in class, but I can never get it right. What am I doing wrong?

Thanks.

4. Jun 22, 2009

### cepheid

Staff Emeritus
The first problem is that you're not seeing what's in the circuit properly.

You can't just calculate like that as though the inductor and the capacitor were in parallel with nothing else in between them. Are you forgetting about R2? The inductor is actually in parallel with (the series combination of R2 and C).

If that confuses you, just slide R2 over until it is directly above C, on the same branch.

You should re-read my first post, where I wrote the correct sequence of steps for the calculation.To understand it (and to write it in words), you must work from the inside out:

1. R2 is series with C...

2. ...ALL of which is in parallel with L

3. ...ALL of which is in series with R1

5. Jun 22, 2009

### cepheid

Staff Emeritus
Oh and by the way, what I actually meant with my question was, what is the impedance of any capacitor (or inductor) in general (as a function of frequency, ω,)

ZC = 1/(iωC)

ZL = iωL

But if you're not familiar with this representation, then don't worry about it. We'll just stick to your phasor notation. It's basically the same thing.

6. Jun 22, 2009

### Petrucciowns

When I do :

C+R2 = 45 ohms 1/45 to the angle of -90 + 1/20 to the angle of positive 90= 1/Sum+ R1

I get 46 ohms instead of the 43.1 I had previously. I am pretty sure the 43.1 is right. What is the answer that you come up with?

7. Jun 22, 2009

### cepheid

Staff Emeritus
My total impedance is a complex number with

amplitude = 43.0563138

and

phase = 48.6214847°

8. Jun 22, 2009

### Petrucciowns

That's what I thought I can't seem to come up with that # can you please post step by step of how you got that # I don't see what I am doing wrong.

9. Jun 22, 2009

### cepheid

Staff Emeritus
10. Jun 22, 2009

### Petrucciowns

Hm those calculations are different then how I learned it ,but I do see they arrive at the same answer. I am going to study them a bit and see if I can remember how I did it in the past.
Thanks for the helps so far.

11. Jun 22, 2009

### Petrucciowns

Thanks, I figured it out I was doing something wrong with the exponents. Now what I am having a problem with is IL and IC. How would I go about those. How do I find the voltage across the parallel section. I have already figured out that it's not source-vr1.

Last edited: Jun 22, 2009
12. Jun 22, 2009

### cepheid

Staff Emeritus
How would you solve the circuit if it were just a resistor network? Remember, what you have here is a bunch of complex impedances, but the same principles apply (provided that you take phase into account). I would start by using Kirchoff's current law at the node that is connected to L, R1, and R2.

13. Jun 22, 2009

### Petrucciowns

I know I have edited this post many times, but at the moment the only thing I have yet to find out is PT.

Last edited: Jun 22, 2009
14. Jun 23, 2009

### cepheid

Staff Emeritus
The total power is just given by:

PT = VTIT

All of these questions just involve combining "phasors" in some algebraic way. Is it possible that there is something about the basic method that you are not understanding that is making it difficult for you to proceed?

15. Jun 23, 2009

### Petrucciowns

See I thought it was that as well, but it does not work in this case. Vt x It = 232w whereas the answer should be 156w.

16. Jun 23, 2009

### Petrucciowns

I know in a series-parallel RLC circuit with one resistor the Power is VT/VR1, but with two resistors with one in series with a parallel section I don't see how to find the total power using those two resistors, but I have a feeling it is calculated with the source and the voltage of both resistors in some combination.

17. Jun 23, 2009

### cepheid

Staff Emeritus
Hi Petrucciowns,

I don't think I can spend too much more time on this thread, so I'll go through a bunch of stuff, just to make sure we're on the same page. It could be that you already know all of this stuff, but I'll go through it anyway. In an AC circuit, quantities are time-varying in a sinusoidal way. For example, the source voltage might be described using the following functional form:

$$v(t) = V_0 \cos(\omega t + \phi)$$​

In this case, $V_0$ describes the amplitude of the oscillation, and $\phi$ describes the initial phase of the oscillation. For mathematical convenience (and this does afford a great deal of convenience), we can describe the oscillating quantity using a complex number $\tilde{V}$ such that:

$$\tilde{V}(t) \equiv V_0 e^{i (\omega t + \phi)}$$​

where this complex quantity has been defined in such a way that its real part is the measurable physical quantity of interest:

$$v(t) = Re[\tilde{V}(t)]$$ ​

This last relation just comes from the definition of a complex exponential. We can rewrite the complex quantity as:

$$\tilde{V}(t) =[V_0 e^{i \phi}]e^{i \omega t}$$​

The quantity in square brackets is a complex number in polar form called a phasor. Like any complex number, a phasor is a vector in the complex plane. Also like any complex number, it can be thought of as two real numbers, the magnitude and the phase (which descibe its length and its angle from the postive real axis respectively). These two numbers are exactly the two numbers that we need to fully-describe the sinusoidal osciallation; the magnitude of the phasor is the amplitude of the oscillation, and the phase of the phasor is the phase of the oscilllation. So we're expressing the information in the most compact possible way. By the way, the time-varying exponential factor in the expression above simply has the effect of causing the phasor to rotate counterclockwise in the complex plane with angular speed $\omega$. As a result, its real part (i.e. the projection of the vector onto the real axis) will vary in length sinusoidally exactly as expected. One final note about phasors: sometimes, instead of writing them the way I did in the square brackets above, they are written as follows:

$$V_0 \angle \phi$$​

Now that all of that is out of the way, we see that in order to calculate the power, we must calculate the power. As far as I have been able to determine, there are two ways to do this. An easy way and a hard way.

The Easy Way

We can use another complex quantity, the apparent power, S, as a convenient formulation of what is going on. S is defined as P + iQ, where P is the true power (the power available to do work on a load), and Q is the reactive power. Reactive power arises from the fact that the voltage and current are not in phase, and are therefore sometimes in the opposite direction, meaning that energy is being stored rather than used up. This energy is just being shuffled back and forth between reactive elements in the circuit, not doing anything useful. I have found out that in terms of the complex voltage and current, the definition of S is:

$$S = \frac{\tilde{V}\tilde{I}^*}{2}$$ ​

Here the * denotes the complex conjugate. Let's say the phase difference between the voltage and current is $\Delta \phi$. Therefore:

$$S = \frac{V_0 e^{i\omega t} I_0 e^{-i(\omega t + \Delta \phi)}}{2} = \frac{V_0 I_0}{2}e^{-i \Delta \phi}$$ ​

We know that the power, P, is the real part of S. Hence,

$$P = \frac{V_0 I_0}{2} \cos(\Delta \phi)$$ ​

Here, $\cos(\Delta \phi)$ is the power factor. The problem is that when you try these with the given numbers, you don't get the right answer:

P = (100 V)(2.32 A)(0.661)/2 = 76.676 W ​

The only thing I can think of is that maybe the given numbers are NOT amplitudes. Sometimes the stated voltage and current values are RMS values. In other words, the phasor is being stated in the form:

$$V_{\textrm{RMS}} \angle \phi$$​

Although RMS values are useful, I think that using them with the phasor notation is confusing and doesn't make sense when you think about what the magnitude of the phasor is supposed to mean. In any case, since these are sinusoids, the RMS values are smaller than the amplitudes by a factor of root 2.

$$P = \frac{V_0 I_0}{2} \cos(\Delta \phi) = \frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos(\Delta \phi) \equiv V_{\textrm{RMS}} I_{\textrm{RMS}} \cos(\Delta \phi)$$ ​

So we are left with the standard result the the real power is the product of the RMS voltage with the RMS current, multiplied by the power factor. So, redoing the calculation, assuming the values given are RMS values already, we get:

P = (100 V)(2.32 A)(0.661) = 153.352 W ​

This is the closest I can get to the answer you have written.

Last edited: Jun 23, 2009
18. Jun 23, 2009

### Petrucciowns

I appreciate all of your help. I wish I could do something to repay the favor.
Daniel

19. Jun 23, 2009

### cepheid

Staff Emeritus
No problem. There were some errors in the coding in my post, but they have been fixed up, so it should be readable now.