Discussion in 'Homework Help' started by weilingkway, Apr 13, 2013.

1. ### weilingkway Thread Starter New Member

Oct 8, 2012
5
1
Hi guys, I really need help on some problems.
The first two attached images are the problems I have to solve (7 problems)
and the third picture is my solution of the first problem in the first picture.
These are take home reviewers, however I don't really understand much of these problems. I only have some ideas on how to solve the problems but I am confused if my answers are actually correct or incorrect.. I'll be posting my attempts on the problems in the pictures... It would be really great if anyone can help me with solving these...

File size:
197.1 KB
Views:
115
File size:
190.9 KB
Views:
96
File size:
83.7 KB
Views:
83
2. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I didn't verify your math, but your work looks correct except for possibly one thing. In calculating the power factor you appear to have ignored the minus sign on the imaginary part of the current. Why?

3. ### weilingkway Thread Starter New Member

Oct 8, 2012
5
1
Truth be told, I forgot about the minus sign, but regardless of the angle having a negative or positive sign the answer would still be the same. It's error on my part :X

4. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Would it be the same? What distinguishes a leading factor from a lagging factor?

5. ### weilingkway Thread Starter New Member

Oct 8, 2012
5
1
The result (regardless of the sign) when you take the cosine of an angle would be the same.

Although AFAIK the one which decides whether the power factor leads or lags is the circuit itself, when the circuit is more inductive (Inductive Reactance is higher than Capacitive Reactance) then the power factor lags, if it is more capacitive then the power factor leads.

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
And that is indicated by whether the phase angle is positive or negative. The magnitude of the power factor does not depend on the sign of the angle, but whether it is leading or lagging does.

So, in this circuit, which is leading which? Current or voltage?

7. ### weilingkway Thread Starter New Member

Oct 8, 2012
5
1
To be honest with you, I don't really know. I think it's the current that leads. For my solution I think it should be leading (Since the + Signs of the imaginary parts of the impedances imply they're capacitors?) ... I'm really confused.

8. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
And that is indicated by whether the phase angle is positive or negative. The magnitude of the power factor does not depend on the sign of the angle, but whether it is leading or lagging does.

So, in this circuit, which is leading which? Current or voltage?

9. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
That's okay, this particular item is confusing and, in fact, is more confusing than it really needs to be because of a convention adopted a century or more ago.

A big part of the problem is that the terminolgy doesn't imply a reference. When we say that the power factor is "leading", what does that mean? Does it mean that the voltage is leading the current or that the current is leading the voltage? This one is fairly easy to keep straight because it makes the most sense, to humans, to use the voltage waveform as our reference. So, in a leading power factor the current is leading the voltage and, in a lagging power factor the current is lagging the voltage.

Now, let's consider an impedance that has an angle θ. If the voltage has no phase angle, then the current will have a phase angle of -θ since I=V/Z.

But is this leading or lagging?

In the time domain, this would mean that the voltage is cos(ωt) and the current is cos(ωt-θ).

But is this leading or lagging?

Assuming that 0<θ<∏, when current will have to wait until t=θ/ω before it reaches the same point that the voltage was at at t=0, thus the current is lagging behind the voltage if the phase angle of the impedance is positive.

So what types of loads have positive phase angles for their impedance? Inductive loads. Therefore, inductive loads have a lagging power factor.

There are a couple of mnemonic that people use to try to keep this stuff straight. One is "ELI the ICE man" which means that E comes before I (i.e., voltage leads current) in an inductive (L) circuit, so ELI. and I comes before E in a capacitive (C) circuit, so ICE. Another is the word CIVIL. CIV => Capacitive means I leads V, VIL => Voltage leads current in an inductor.

In both cases, you need to then remember that E leads I is the same as I lags E.

So, with all of that in mind, is your power factor leading or lagging?

10. ### weilingkway Thread Starter New Member

Oct 8, 2012
5
1
It's Lagging. My guess of the power factor in #1 being "leading" is in fact wrong.

My initial assumption was that those positive reactances were from capacitors and the negative ones were from inductors. Apparently this is also wrong...

I tried solving #2 as well and it turns out (As far as I understand it) the convention that "if Z = R + jX" then X = Capacitive Reactance and "if Z = R - jX" then X = Inductive Reactance is actually incorrect (I don't really know but it doesn't seem to be that way for me, I'm still kind of lost with the convention... The only thing I understand is that the signs depend on which type of reactance is greater...)

I'm getting more and more confused as to what must be the first assumption when given an impedance with positive or negative reactance... Is it the inductive reactance which is positive or the capacitive reactance? D:

Anyway, here's my solution for #2.. If you could please check it, I hope I got this one right..

• ###### 4.jpg
File size:
242.2 KB
Views:
38
John Ramelb likes this.
11. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
A positive reactance is inductive and a negative reactance is capacitive.

This will make a huge amount of sense to you when and if you get introduced to Transform Methods using Fourier and Laplace transforms.

Basically,

Z_R = R
Z_C = 1/(jωC) = -j/(ωC)
Z_L = jωL

See where the minus sign comes from?

Now, why the jωL is in the numerator and the jωC is in the denominator has to do with taking the Laplace (or Fourier) transform of the differential equations that describe the voltage/current relationships for an R, an L, and a C.