# Stuck and confused on a question about RLC circuits with no given impedences.

Discussion in 'Homework Help' started by matt_t, Aug 19, 2012.

1. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
Hi everyone I was wondering if someone could help me with the following question and explain their workings. I know how to work out RLC circuits but i'm confused and have no idea where to start with this one:

Two single-phase motor are connected in parallel to a 240 Volt, 60 HZ supply. Motor 1 draws a current of 20 Amps at a power factor of 0.67. Motor 2 draws a current of 40 Amps at a power factor of 0.79. If a capacitor is used to correct the power factor to 0.93 the current through the capacitor would be _____ Amps.

2. ### WBahn Moderator

Mar 31, 2012
21,551
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First, model each motor as an inductor in series with a resistor. What values do you get for the components?

3. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
Hi, thats the thing I only have the information which is in the question. Which is the motors currents which is resistive and inductive combined. If I add 20 /_ 47.93°
+ 40 /_ 37.81° then I have the total circuit current 59.79 /_ 41.18° am I correct?

4. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
I didn't get quite what you got, but it looks like you have the right idea. I got 60.12A @ 41.25deg, but I wrote down the intermediate results on paper.

So what is the next step?

Last edited: Aug 19, 2012
5. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
I = 240 /_ 0° / 40 /_ 30°
= 6 /_ -30°

PF= cos -30° = 0.866

6. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
You can forget the little problem I posed. I didn't read your prior post closely enough at first to see that you had already done what I was trying to lead you to.

Q1) So what is the equivalent impedance of the two motors combined?

Q2) What does the impedance angle need to be in order to result in a power factor of 93%?

Q3) Given this impedance and a capacitor, C, placed in parallel with it, what does the value of C need to be in order to achieve that impedance angle?

Q4) What current will be flowing in the cap?

7. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
1) 240 /_ 0 divided by 59.79 /_ 41.18 = 4.01 /_ -41.18 ohms
2) -cos 0.93 = 21.57 degrees
3) Not sure how I go with this one, because I don't have the Xc only the total Z? 1/ 2piex60xXc?

8. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
Now, before we go further, we need to ask whether we have the sign on the angles correct.

Remember, cos(theta) = cos(-theta).

What sign do you expect for an impedance that represents a motor?

If the load is only corrected to 93%, do you think the net result is capacitive or inductive?

Getting back to the problem. You have an impedance, Z1, that represents the motors. You are going to put a capacitor, C, in parallel with that Z1. What will the resulting impedance be? What will it's angle be? What value of C will make that angle what you want it to be?

9. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
I'm not sure what you quite mean about the signs on the angles? you mean + -?

The net result will still be inductive.

The corrected angle will be 21.57 degrees.

1/ 2xpiex60x4.01 = 38.29F? im lost.

10. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
1 /2xpiex60x4.01 = 66mF

11. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
Let's say that I have Z4=30Ω @ +20deg and Z5= 30Ω @ -20deg. What will the power factor be for each?

With the answer to this in mind, given a power factor, do you know whether the impedance angle is positive or negative?

What is the difference between Z4 and Z5 (i.e., why does one have a positive impedance angle and the other a negative one)?

Why do you say that? I'm not saying you're wrong (and I'm not saying you're right), just want you to articulate your reasoning.

So is that inductive or capacitive?

It's 'pi' (the Greek letter), not 'pie' (the dessert).

4.01Ω is the magnitude of Z1, not the magnitude of the capacitive reactance of the compensating capacitor.

Take a step back and consider the following problem:

I have an inductor, L, a capacitor, C, and a resistor, R. I take the inductor and the resistor and put them in series to form Z1. I now put the capacitor in parallel with Z1. What is the effective impedance of all three components? What is the impedance angle?

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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In the absence of any requirement to work out the actual load impedances I'm puzzled as to why one would then go to all that trouble. The question asks only what the resulting capacitor current would be with the power factor corrected to 0.93 [presumably lagging].

The correct uncompensated load current would be

$20\angle{-47.93^o}+40\angle{-37.81^o}=59.79\angle{-41.17^o}=45-j39.37 \ Amps$

The compensating capacitor current then simply brings the total source current to

Itot = Ireal/(compensated pf)=45/0.93=48.39 Amps

The resulting (compensated) source reactive current is then
Ireactive = Itot*sin(acos(0.93)) = ??? Amps

The original source reactive current was 39.37 Amps. The capacitor current is then the difference between the original and compensated source reactive current values.

13. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
For Z4=30Ω @ +20deg and Z5= 30Ω @ -20deg the power factor will be cos 20 = 0.939 and cos -20 = 0.939.

With just the power factor I can't tell whether its positive or negative or (inductive, capacitive).

The difference is that 0-180 deg is inductive and 180-360 is capacitive and -20 is the same as 340 deg.

I'm saying the final power factor will be inductive as I know the motors are basically inductors in series with a resistor but the more I think about it the more confused i'm getting. What if they used a larger than needed capacitor and the circuit became more capacitive.

14. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
You are making progress.

For the motors themselves, before you put the capacitor in, you would expect the load to be inductive. Hence, you would expect the impedance to have a positive angle. But you came up with a negative angle. Why? Because you have a positive angle for the current (and zero angle for the voltage). Why? Because you got the angle for the current by taking the arccos of the power factor. But, as seen above, the resulting angle could be positive or negative, so you can't just blindly take the positive one. In this case, you need to take the negative one. So your current is 59.79A @ -41.18° and you impedance then has an angle of +41.18°.

You are correct that you don't know whether the final power factor of 93% is inductive or reactive and there really isn't enough information to make a firm decision. So solve it both ways and provide both answers.

While t_n_k is correct that you can find the current without ever finding the capacitance, I'm trying to keep you going this route because I think we have exposed a weakness in your understanding of circuits. So I am trying to overcome and correct the weakness in addition to getting an answer to the problem (and solving it directly will serve as a useful check on the results).

So, if you are willing, let's push ahead.

You have Z1, the effective impedance of the two motors in parallel as

$
Z_1\;=\;R+jX_L\;(=4.01 \Omega \angle 41.18^o)
$

You have a capacitor, which has an impedance of

$
Z_C\;=\;jX_C\;(X_C = -\frac{1}{2\pi f C})
$

In terms of R, X_L and X_C, what is the impedance of Z_1 and Z_C if they are in parallel?

15. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
Hi, I dont understand why you have negative impedance angles after reading this from here :http://www.allaboutcircuits.com/vol_2/chpt_5/2.html

16. ### matt_t Thread Starter Member

Aug 18, 2012
30
0

I dont know how I can find the impedance of Zc as I have no value for C or Xc all I have is a pf of 0.93 or 21.57 deg

17. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
So to confirm the total impedance of the circuit, I worked out the combined Z being:

240 /_ 0 divided by 59.79/_ -41.18 = 4.01 /_ 41.18 deg or 3.01 +j2.64 so I have a total R of 3.01Ω and a total XL of 2.64Ω?

18. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
Actually, you had the negative impedance angle clear back in Post #7 and I am maintaining that it should be positive. But I'm not completely sure which impedance angle yoiu are referring to above.

But, regardless of that, the point you are making is worth considering because this is one of those areas in which there are two camps and you can find plenty of reference material supporting both views. I generally lump the two camps into a "technician" view and an "engineering" view, but the difference is primarily due to the level of mathematical underpinning that is used. The "technician" view is principally oriented to explanations that do not rely on any understanding of complex numbers or transform methods in circuit analysis, so you end up seeing a lot of formulas thrown about without any derivations to go along with them. Various conventions, particularly sign conventions, then grew up out of convenience from certain perspectives independent of what made the most sense from a mathematical perspective. The "engineering" view is much more tightly integrated with the math so that even when material is introduced prior to the student having the mathematical background to understand where some of it comes from, it is known and expected that the student will eventually reach that point and have to deal with it; therefore, much more care is taken to make the sign conventions match the mathematical foundations.

From an engineering perspective impedance is a complex quantity that relates the ratio of the Fourier transform of the voltage across a linear component to the Fourier transform of the current through that same component (actually, its more general than that and involves the Laplace transform, of which the Fourier transform is a special case, but this is more than adequate for our purposes here). Namely:

$
V(j\omega)\;\equiv\;\mathcal{F}\{v(t)\}
\;
I(j\omega)\;\equiv\;\mathcal{F}\{i(t)\}
\;
Z(j\omega)\;\equiv\;\frac{V(j\omega)}{I(j\omega)}
$

The impedance can be written with a real and an imaginary part, namely:

$
Z\;=\;R\;+\;jX
$

Where
Z is the complex impedance
R is the resistance
X is the reactance

Mathematically, this is:

$
R\;\equiv\; \mathcal{Re}\{Z\}
X\;\equiv\; \mathcal{Im}\{Z\}
$

We have three types of linear components we are usually concerned with, resistors, capacitors, and inductors. The constitutive relations for these three are:

$
R:\; v(t)\;\equiv\;R \cdot i(t)
\;
C:\; Q \equiv C \cdot V
C:\; \frac{dQ}{dt}\;=\;i(t)\;=\;C \cdot \frac{dv(t)}{dt}
\;
L:\; v(t)\;\equiv\;L \cdot \frac{di(t)}{dt}
$

If we take the Fourier transform of these, we get:

$
R:\; V(j\omega)\;=\;R \cdot I(j\omega)
\;
C:\; I(j \omega)\;=\;C \cdot (j\omega V(j\omega))
\;
L:\; V(j\omega)\;=\;L \cdot (j\omega I(j\omega))
$

The jw factors that appear on the RHS for the capacitor and inductor are a consequence of taking the Fourier transform. Basically, the Fourier transform of the derivative of a function is jw times the Fourier transform of the function itself. Don't worry if it doesn't make sense to you (unless you have had Fourier transforms, of course). This is the only piece of mathematical magic that is involved here.

We can now find the impedance of each of these components:

$
Z_R(j\omega)\;=\;R
\;
Z_C(j \omega)\;=\;\frac{1}{j\omega C}\;=\;\frac{j}{j^2\omega C}\;=\;j \left( \frac{-1}{\omega C} \right)
\;
Z_L(j\omega)\;=\;j\omega L
$

Written in full-blown rectangular form, these are:

$
Z_R(j\omega)\;=\;R+j0
\;
Z_C(j \omega)\;=\;0+j \left( \frac{-1}{\omega C} \right)
\;
Z_L(j\omega)\;=\;0+j\omega L
$

Noting that

$
\omega\;=\;2\pi f
$

the resistance and reactance for each component is therefore:

$
\text{RESISTOR}
R_R\;=\;R
X_R\;=\;0
\;
\text{CAPACITOR}
R_C\;=\;0
X_C\;=\;\frac{-1}{\omega C}\;=\;\frac{-1}{2\pi f C}
\;
\text{INDUCTOR}
R_L\;=\;0
X_L\;=\;\omega L\;=\;2\pi f L
$

Notice that at no point did we have special rules for any of these three components; they were treated identically and the same operations and definitions were applied to each. You don't have to track separately which reactances are for inductors and which are for capacitors and then be sure to subtract the capacitive reactances from the inductive reactances and then make a note of whether the result is capacitive or inductive. A reactance is a reactance. It is a signed quantity and you can let the sign work its way through the math the same way you would any signed quantity.

A review of my text books shows that the honors are split but with the engineering view being the more prevalent. Interestingly, a couple of the texts (including my Transform Methods text) that take the technician view almost immediately contradict themselves because they have to treat the reactance of an arbitrary component as being a signed quantity but then want to have the capacitive reactance be positive, even though the reactance of an arbitrary component that just happens to be a capacitor has to be negative. Personally, I find that entire path to be needlessly confusing and ripe for mistakes and miscommunication. Mych better to have a simple, consistent definition of things and let the chips fall where they may.

19. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
I agree with the idea that things should be simple and consistent. I still don't feel like I have a clear explanation on how to answer the original question, so basically the total current of the circuit (45 -j39.37) is divided by 0.93 to give the corrected new total circuit current of 48.39A. To find the reactive current I do 48.39 x sin 21.57 = 17.94A. Original reactive current was 39.37 so 39.37-17.94= 21.43A through the new capacitor?

20. ### WBahn Moderator

Mar 31, 2012
21,551
6,218
You treat C (or Xc) as the unknown that you are trying to solve for.

Since you have an alternative way to do it, and since you have been struggling for a while, but putting forth the effort, I'll set it up for you.

You have two impedances, Z1 and Zc, that can be expressed as follows:

$
Z_1\;=\;R+jX_L
Z_C\;=\;jX_C
$

$
Z_1 \;=\; 4.014\Omega \angle 41.18^o
\;
R \;=\; Z_1 \cos(41.18^o) = 3.021\Omega
X_L \;=\; Z_1 \sin(41.18^o) = 2.643\Omega
$

These two are in parallel, so the effective impedance, Z, is therefore:

$
Z \;=\; \frac{Z_1Z_C}{Z_1+Z_C}
\;
Z \;=\; \frac{(R+jX_L)(jX_C)}{(R+jX_L)+(jX_C)}
\;
Z \;=\; \frac{-X_CX_L+jRX_C}{R+j(X_L+X_C)}
$

We can normalize this by multiplying top and bottom by the complex conjugate of the bottom:

$
Z \;=\; \frac{-X_CX_L+jRX_C}{R+j(X_L+X_C)} \;\cdot\; \frac{R-j(X_L+X_C)}{R-j(X_L+X_C)}
\;
Z \;=\; \frac{(-RX_CX_L + RX_C(X_L+X_C))+j(X_CX_L(X_L+X_C) + R^2X_C)}{R^2+(X_L+X_C)^2}
\;
Z \;=\; \frac{RX_C^2+jX_C(R^2+X_L(X_L+X_C)}{R^2+(X_L+X_C)^2}
$

While this looks ugly, it's actually not too bad given what we want to do with it.

We want to find the value of Xc that will result in the angle of this impedance being such that the power factor will be 93%.

If the power factor is 93%, the impedance angle as to be either +arccos(0.93) or -arccos(0.93), meaning that the angle as to be ±21.57°.

Since the Z above is normalized (only the numerator has an imaginary component), the angle of Z is the angle of the denominator, so we know the tangent of the impedance angle is:

$
\tan(\theta) \;=\; \frac{X_C(R^2+X_L(X_L+X_C))}{RX_C^2}
\;
\tan(\theta) \;=\; \frac{R^2+X_L(X_L+X_C)}{RX_C}
$

But we know what we want the angle to be, so we know what we want the tangent of the angle to be:

$
\theta \;=\; \pm 21.57^o
\alpha \;=\; \tan(\pm\theta) \;=\; \pm 0.3952
\;
\alpha \;=\; \frac{R^2+X_L(X_L+X_C)}{RX_C}
\;
\alpha RX_C \;=\; R^2+X_L(X_L+X_C)
$

Now solve for Xc:

$
\alpha RX_C \;=\; R^2+X_L(X_L+X_C)
\;
\alpha RX_C - X_LX_C\;=\; R^2+X_L^2
\;
X_C(\alpha R - X_L)\;=\; R^2+X_L^2
\;
X_C\;=\; \frac{R^2+X_L^2}{\alpha R - X_L}
$

Note that the numerator is just the square of the magnitude of the impedance of Z1. Also, we expect the reactance to be negative (since it is a capacitor) and so we will pull that out of the denominator:

$
X_C\;=\; -\frac{|Z_1|^2}{X_L-\alpha R}
$

Plugging in numbers, we have:

$
X_C\;=\; -\frac{|4.014\Omega |^2}{2.643 \Omega \; \mp \; 0.3952 \cdot 3.021 \Omega}
$

This will then yield two capacitive reactances that will yield an overall power factor of 93%; one will correspond to the overall circuit still being inductive and the other will correspond to the overall circuit now being capacitive.

Last edited: Aug 19, 2012