**what is the average power pavg supplied by the voltage source?** This is a topic that many people are looking for. **bluevelvetrestaurant.com** is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, ** bluevelvetrestaurant.com ** would like to introduce to you **Electrical Engineering: Ch 12 AC Power (10 of 38)What is Average Power Supplied and Absorbed?(-26.5)**. Following along are instructions in the video below:

To electron line. The following problem here is a simple circuit. Where we have a a power supply.

We have a resistor and we have a capacitor and were to find the power supplied by the power supply and the power consumed or absorbed by the resistor and the capacitor. So here we have the equation to find the average power. Now what we need to do in each case is we need to know the current through the circuit through each of the components.

And we need to know the voltage across each of the components and of course. We already know the voltage across the power supply. We still need to find the voltage across a resistor and across the capacitor.

So lets start with finding the current first what we can say here is that the current. I is equal to the voltage divided by the total impedance and so here. What we could do is we have the impedances of a sister and a capacitor combined.

So in this case. I will be equal to 5 volts with a phase angle of 30 degrees. Divided by that would be 4 minus j2.

Now to make out the vision. We probably want to change that to the magnitude and phase angle format. So this is equal to 5 for the phase angle of 30 degrees divided by okay the magnitude of that that would be 16 plus.

4. Thats the square root of 20 and that would be 4 point 4 7. And a phase angle of so this is a negative.

Here so we have 2 divided by. 4 thats 05. In the negative direction.

Therell be twenty six point five six five degrees. Okay and if we divide one by the other lets see here. 26.

That becomes negative. We add up to 30. So thats a phase angle of three point four three degrees.

And the magnitude here would be five divided by four point four seven. We get one point one one nine one point one one nine for the magnitude all right well actually when i dont round it off my my better by writing eight not that it makes a lot of difference in the third decimal place. But thats probably a little bit more correct okay so now we have the current in magnitude and phase phase angle format.

So now we need two voltage across each of the three components well we know the voltage across here how about the voltage across there so that case the voltage across the resistor is going to be equal to the current times. The resistance. We know the current its one point one one eight with a phase angle of three point four three degrees and were going to multiply that times resistance.

Which is four with the phase angle of zero degrees. And lets see here nup times four that gives us about four point four seven four point four seven volts and with a phase angle of three point four three degrees. All right so thats the voltage across the resistor now the voltage across the inductor.

So not injective a capacitor the voltage across the capacitor is a current times x sub c. So in this case that would be one point one one eight with a phase angle of three point four three degrees. And were multiple at a times two with a phase angle of minus ninety degrees divide.

That by lets see here divided by 4 and multiply it times. 2 all. Right.

There that gives us a magnitude of 22. 4. And the phase angle would be this minus.

That that gives us a eighty six point five seven degrees. And that would be a minus and so that would be both of course in terms of volts. So now we have the potential to voltage across each of the two components now were ready to find the power supply that absorbed by the various components.

So first lets take the power supplied by the source. So the power.

Supplied by the voltage source is equal to 1 2. Times. Imax imax.

Thats the same for all of them. That would be one. Point.

One. One eight multiplied times. V.

Max. That would be five and times. The cosine of the difference in the phase angle.

Thatll be three point four three for the current thats degrees of course. 30. The phase angle for the voltage.

And lets see here that gives us three point four three 30 take the cosine of that multiply that times five. Times one point one one eight and times 05. Equals and thats two point five.

And that would be watts. So that would be the power supplied by the power supply. So now were going to take now.

Were going to try to find the the power absorbed or yeah absorbed by the resistor power by the resistor is equal to one half times. The current one point one one eight the voltage across the resistor four point four seven and that would be times. A cosine of three point four three degrees.

And that would be again three point four three degrees. Because we would not expect to find any phase angle across the resistor and so therefore that would be the cosine of zero degrees. Which is one so now we have four point four seven times.

One point one one eight times one five and again two point. Five so its equal to 25 watts. But in this.

Case this is absorbed and this is supplied the. Power. Supply supplies.

25. Watts and the resistor. Absorbs 25.

Watts. What about the capacitor. Well lets find out the power across the capacitor is equal to 1.

2 times the current. 111. 8.

Times the voltage across the. Capacitor which is 22. 4.

Times. The cosine of the difference between the phase angle of the current which is 343 degrees minus a minus eighty six point five seven degrees like that and notice that this would be equal to one half times. One point one one eight times two point two four times.

The cosine of ninety degrees and of course the cosine of ninety degrees. Thats equal to zero so zero times. Anything else that would be equal to zero watts.

So the capacitor itself does not consume any power that means that all the power supplied by the power supply is consumed by the resistor and none of it is absorbed by the capacitor thats how its done. .

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