# ac analysis: maximum power transfer

#### PG1995

Joined Apr 15, 2011
832
Hi

Regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Regards
PG
Actually, if XL=-Xth, then RL=Rth. If you substitute eqn. 17 into eqn. 18 then it automatically follows.

#### PG1995

Joined Apr 15, 2011
832
Thank you.

I'm sorry I thought the author was simply adding them up... It's not that I haven't asked such dumb questions before!

Best wishes
PG

#### steveb

Joined Jul 3, 2008
2,436
Thank you.

I'm sorry ... It's not that I haven't asked such dumb questions before!
You're welcome.

Actually, you haven't asked any question as dumb as that. But, everyone has to own a dumbest question, and there are much worse ones to own than that.

I'm too embarrassed to tell you mine!

Anyway, mistakes and oversights are committed the best of us, and sometimes more often by the best of us, so don't be sorry.

#### PG1995

Joined Apr 15, 2011
832
Actually, you haven't asked any question as dumb as that.

You are so nice. That's true that you are not infallible despite being very knowledgeable, as Mrs. Steve thinks you mentioned once, but one thing is true that you are one of the nicest persons I have come across. Actually this forum is full of so many nice members who are willing to help without ridiculing you or making fun of your questions.

With best wishes
PG

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#### tony caney

Joined Jan 13, 2012
1
The RTH is the Thevenin equivalent cct for maximum power transfer

#### PG1995

Joined Apr 15, 2011
832
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PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Regards
PG
Double check your complex math calculation. Magnitude squared is indeed found by multiplying by the complex conjugate. But, the answer always comes out to a real number, if you do the math without errors.

(2+7i)(2-7i)=4+14i-14i+49=53

#### PG1995

Joined Apr 15, 2011
832
Hi

Regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Regards
PG
It's not clear to me how you are trying to calculate the power in the capacitor. If you trying to calculate average real power, then you must realize that the voltage and current are always 90 degrees out of phase in either a coil or a capacitor. If you do out the integral for average power over one period, you'll see that the 90 degree phase shift results in the product of voltage and current forming a symmetrical function above and below the time axis. Hence, the integral is zero. So, reactive components (coils and caps) have only reactive power and no real power. Reactive power is just a case of power flowing out and then flowing back in over the course of a period.

The resistor power formula is derived in the same way, but since current and voltage are in phase, the product of the two is always greater than zero, and hence the integral can't be zero unless both voltage and current are zero (EDIT: I meant to say "either" voltage or current is zero).

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#### PG1995

Joined Apr 15, 2011
832

First I found the current through the capacitor, Ic. Then, I found the voltage across the capacitor, which is same as across the 4 ohm resistor, hence V_4Ω. Then, I used the formula: (1/2)Vm*Im*cos(θv-θi), Vm and Im are magnitudes of voltage and current. In another question I just solved, I ended up correct answer for the power absorbed by an inductor, i.e. 0W. In this solved problem I ended up with the correct answer for the power absorbed by a capacitor.

Please also help me with the Q1 in the attachment. Thanks.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
OK, I understand your approach. But, look at the answer you got in the first attachment. The capacitor voltage and capacitor current you calculated were not 90 degrees out of phase. This indicates that you made a math error somewhere. I would recommend that you just start over on a fresh piece of paper and carefully derive it out. If somehow you dont get Ic and Vc 90 out of phase, post that new analysis and I'll see if I can spot where the error is.

#### PG1995

Joined Apr 15, 2011
832
Thank you, Steve.

Best regards
PG

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#### steveb

Joined Jul 3, 2008
2,436

Best regards
PG
So, what are questions Q1 and Q2? I think you pointed to the wrong page.

#### PG1995

Joined Apr 15, 2011
832
Hi

Regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
Hi

Regards
PG
For AC, direction is ambiguous. With DC, the idea of direction is clear. The electrons either always flow one way or the other way. With AC, electrons flow in both directions. In some sense direction is meaningless, or at least it is relative to something else. However, we can develop the notion of abstract direction by using the angle as the definition of direction. Here we demand that the magnitude is positive, and the resulting angle is the direction. If magnitude is negative, then make it positive and add or subtract 180 degrees from the angle. That angle is the direction.

Rather than thinking of direction, I prefer to think of phase difference or phase shift, but the notion of direction is an accepted terminology and viewpoint with AC signals, and more particularly, the idea of a space vector, which is viewed as having a real spatial direction (even though it doesn't) is used with 3-phase machines.

#### PG1995

Joined Apr 15, 2011
832
Thanks a lot, Steve.

It seems I wasn't clear enough in my previous post because your reply apparently answers something else, or it's just me. Anyway, I have made the queries in a different way so that you can hopefully easily understand what is troubling me. Please help me. Thank you.

Best wishes
PG

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#### steveb

Joined Jul 3, 2008
2,436
I feel my answer above basically answers the questions, but we can come at it differently if it's too confusing that way.

For Q1, the current in the 4 ohm resistor is I2-I1 because of the way you defined the current direction through I4. You could choose to say that I4 is in the other direction and you would get I1-I2, as you say. The math would work out fine either way. I'm not sure why this would bother you, but remember that you don't directly subtract magintudes. Instead you convert to rectangular representation of complex numbers and then subtract. The math just takes care of itself.

For Q2, there is no incorrect direction. It works either way. With DC, we can do it either way and if we get a minus sign, then we know the real current is in the opposite direction. The negative sign takes care of it, no problem. With AC, what is the basis to say you even chose the wrong direction? Current flows both ways.

#### PG1995

Joined Apr 15, 2011
832
Thank you, Steve.