#### lisa3412

Joined Sep 6, 2014
37
Hello,

I have uploaded an image of the Question.

I have worked out the answer and it comes out to be 40.96 @ angle -36.87 Degrees.

Thank you

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#### t_n_k

Joined Mar 6, 2009
5,455
Don't believe you are correct.

#### t_n_k

Joined Mar 6, 2009
5,455
Might be worth checking your work Nandu.
Consider the Thevenin impedance looking back from the load into the internal network.
Plus it's generally not considered to be "good form" to give answers on the homework forum unless the OP has shown some working and made a reasonable effort to understand & solve the problem.

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#### MrAl

Joined Jun 17, 2014
8,372
Hi,

Yeah that's a good idea tnk. There's no proof that any answer thus far is correct yet either

#### lisa3412

Joined Sep 6, 2014
37
Hey Guys,
So, this is how I worked it out:

ZTH of the system is 4+j3-j6=4-j3ohms

Is this logic correct?

#### studiot

Joined Nov 9, 2007
4,998
Not quite as easy as this I'm afraid, lisa.

Remember that ab is measured across the capacitor alone.
To calculate the Thevenin equivalent impedance, looking in from ab you need to short circuit the voltage source.

Thus the combination of R, L and C is more complicated than just series.

Can you see what this is?

#### lisa3412

Joined Sep 6, 2014
37
Yes I did look at it from that approach. I short circuited the Voltage and summed up the Impedances in Series.
Can you tell me how to solve this then? If it is not simple addition?

#### studiot

Joined Nov 9, 2007
4,998
Did you draw a diagram?

#### lisa3412

Joined Sep 6, 2014
37
No. I went wrong here. I took R+L+C in series. I should draw figures in the future. Thank you . I shall post my answer in a bit. Please let me know if it is right.

#### studiot

Joined Nov 9, 2007
4,998
You can obtain the current flowing with ab open circuit by summing the series combination ot R,L and C.

This will then enable you to calculate the Thevenin voltage across C and ab.

Sorry I switched the R and L round in my sketch, but it makes no difference to the outcome.

#### lisa3412

Joined Sep 6, 2014
37
but thevenin voltage aint required in this problem right?

#### studiot

Joined Nov 9, 2007
4,998
When you replace the voltage source with the thevenin equivalent impedance ( in series with the load), what voltage will you drive it with?

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#### lisa3412

Joined Sep 6, 2014
37
Ok. So that means when we calculate P_load=V*I*Power factor, We take V=Vth. NOT V=16@ angle zero. Am i right?

#### studiot

Joined Nov 9, 2007
4,998
What does Thevenin's Theorem say?

#### lisa3412

Joined Sep 6, 2014
37
Thevenin's theorem basically is replacing a given network as viewed from two o/p terminals by a single voltage source with series Impedance right?

#### lisa3412

Joined Sep 6, 2014
37
If you look at the original question, we have to find out P_load max right? P=I^2*R
Here R=4 ohms
I= Vs/(Zth)*

Is this right?

Or is it I= Vth/Zth where Vth= voltage across C

#### studiot

Joined Nov 9, 2007
4,998
I have run through the method of the problem in a way that I hope will also help you understand Thevenin better.

Vth is the voltage across the cut section ab (which yes is the voltage across the capacitor)

Yes, Nandu had a point (the power is in the resistive or real part of Zth) but you can see why tnk asked him to check his Zth

#### lisa3412

Joined Sep 6, 2014
37
Hey guys,

I really appreciate your help. Yes, I have understood the concept and arrived on the answer of 16.06W. Thank you

#### studiot

Joined Nov 9, 2007
4,998
Yes 16 watts is correct. I expect the 0.06 is due to rounding error in the calculations.

#### MrAl

Joined Jun 17, 2014
8,372
Hi,

Another interesting idea here is since there may not be anybody else around to tell you that "Yes, 16 watts is correct", or "No, it's not correct", you should look for ways to prove your result by yourself. Sometimes you wont want to go through a formal proof, but at least you can get an idea if you probably have the right result by doing a simple test or two. For many problems like this you may be able to come up with a second method for doing it and thus come out with the same answer if it is right, but other times it is good enough to just try a couple little experiments.

For this problem, the idea is that the power in the real part of the load has to be a maximum when the load impedance is selected properly, so once you come up with a result (like 16 watts) you can test that idea by varying the load impedance just a little bit (and you must have calculated the load impedance already anyway).

So the pseudo proof would go something like this...
You calculate the load impedance and you get a certain value a+b*j, and when you connect that as the load you get a certain power dissipation in the resistive (real) part of the load, and in this case it is 16 watts. But you cant be sure if that is right or not. And what if everybody who replied made the same mistake you made, then you would still not have the right result. So to test it, vary the load a little bit. It may also help to convert the load into an actual R and L or an R and C depending on what kind of load it came out to be, as that injects a little more realism into the problem, but that is up to you. In any case, if you vary the real part (resistor) or the imaginary part (inductor or capacitor) you should see a DECREASE in the resulting power in the real part of the load. If you do not see this behavior then you dont have the right result yet.
In this case you came up with 16 watts. If you change either the R or the L in the load, you will see the power get lower (like possibly 15.8 watts) no matter what you do to the load. If he load power goes higher though (16.5 watts for example) then you probably dont have the right load yet.
We know that 16 watts is correct, but before there is any confirmation of that it might still be unsure so the best thing to do is simply test it.

This works with a lot of problems, but the kinds of tests you have to do wont always be the same.