# average power in impedance

#### e-learner

Joined Apr 25, 2015
30
here is the problem:
given the time-domain voltage v=4cos?(πt/6) V, find the average power that result when the corresponding phasor voltage V=4∠0° V is applied across impedance Z= 2∠60° Ω.
My question is : As average power is present only inthe resistive component, so I need to calculate for Re|Z|=1 Ω and Veff=2√2 rms. and then the average power P= 8 watt(using P=Veff^2/R)
but as solved in the example , V/Z=2∠-60° A and P= 1/2(4)(2) cos 60°= 2 watt.
what's wrong with the first method?

#### WBahn

Joined Mar 31, 2012
29,175
Is the effective voltage really the voltage that appears across the resistive part of the impedance?

#### e-learner

Joined Apr 25, 2015
30
Is the effective voltage really the voltage that appears across the resistive part of the impedance?
Isn't it? Because, its mentioned that the voltage is applied across the impedance.
There were other examples in the book , for e.g.: calculate aperage power delivered to the impedance of 6∠25°Ω bythe current I= 2+j5. and the fact that average power appears across resistive part , was used to solve it. I don't know why can't apply the same fact here.

#### WBahn

Joined Mar 31, 2012
29,175
Isn't it? Because, its mentioned that the voltage is applied across the impedance.
There were other examples in the book , for e.g.: calculate aperage power delivered to the impedance of 6∠25°Ω bythe current I= 2+j5. and the fact that average power appears across resistive part , was used to solve it. I don't know why can't apply the same fact here.
Yes, across the impedance. Not across just the resistive portion of the impedance. When you split the impedance apart you are effectively creating a voltage divider with two series components formed from the resistive and reactive parts. In the book example they were using a current and the same current flows in both components.

#### MrAl

Joined Jun 17, 2014
10,620
Isn't it? Because, its mentioned that the voltage is applied across the impedance.
There were other examples in the book , for e.g.: calculate aperage power delivered to the impedance of 6∠25°Ω bythe current I= 2+j5. and the fact that average power appears across resistive part , was used to solve it. I don't know why can't apply the same fact here.
Hi,

Are you sure you are not making some simple mistake?

You can either use 4/sqrt(2) as the source or Vr/sqrt(2) once you know the peak voltage across the resistor.
So if you use 4 as the source, then you have to use Vr/sqrt(2) for the resistor, but if you use 4/sqrt(2) then you cant use Vr/sqrt(2) for the resistor because you've already accounted for the fact that the voltage was given as the peak.
However, that doesnt change the fact that the CURRENT is also going to be different. For example, if the peak voltage was 3v then 3/sqrt(2) is the rms voltage, and then 3/(sqrt2) is the rms current (R=1), so
[3/sqrt(2)]*[3/sqrt(2)] is equal to 9/2 which is equal to 4.5, but that's just a different example of the application.

So either do it with the source or with the voltage across the resistor, but be sure to calculate the RMS current too.

If you still have trouble we'll have to go over the whole solution i guess. Not a problem, but try once more first. If you like you can check your answer also by doing it in the time domain.

#### e-learner

Joined Apr 25, 2015
30
Hi,

Are you sure you are not making some simple mistake?

You can either use 4/sqrt(2) as the source or Vr/sqrt(2) once you know the peak voltage across the resistor.
So if you use 4 as the source, then you have to use Vr/sqrt(2) for the resistor, but if you use 4/sqrt(2) then you cant use Vr/sqrt(2) for the resistor because you've already accounted for the fact that the voltage was given as the peak.
However, that doesnt change the fact that the CURRENT is also going to be different. For example, if the peak voltage was 3v then 3/sqrt(2) is the rms voltage, and then 3/(sqrt2) is the rms current (R=1), so
[3/sqrt(2)]*[3/sqrt(2)] is equal to 9/2 which is equal to 4.5, but that's just a different example of the application.

So either do it with the source or with the voltage across the resistor, but be sure to calculate the RMS current too.

If you still have trouble we'll have to go over the whole solution i guess. Not a problem, but try once more first. If you like you can check your answer also by doing it in the time domain.
I understand my mistake that I need to calculate V across resistor in order to use P= Veff^2/R.
I understood your point also, but I didn't make that mistake as I was not using P=Veff X Ieff formula and in the book example '1/2' was accounting for the rms values of voltage and current.

But thanks, I 'll keep your point in mind.

#### MrAl

Joined Jun 17, 2014
10,620
Hello again,

I was throwing some ideas out there. I got the book result so i was suggesting some things that might have went wrong.

So what was your mistake again, and do you get the right result now?

It is also interesting that when you integrate sin(wt)^2 over one period and divided by the period you get 1/2.

Last edited:

#### WBahn

Joined Mar 31, 2012
29,175
Hello again,

I was throwing some ideas out there. I got the book result so i was suggesting some things that might have went wrong.

So what was your mistake again, and do you get the right result now?

It is also interesting that when you integrate sin(wt)^2 over one period you get 1/2.
You don't get 1/2, you get pi.

Now, the AVERAGE value of sin(wt)^2 over one period is 1/2.

That latter observation can be seen by noting that

sin(wt)^2 + cos(wt)^2 = 1

combined with the observation that, over one period, the integral of sin(wt)^2 = the integral of cos(wt)^2.

Thus,

$$y_{avg} \: = \: \frac{1}{2 \pi} \int_0^{2 \pi} \sin^2 \( \Theta$$ d \Theta
\;
y_{avg} \: = \: $$\frac{1}{2 \pi}$$ $$\frac{1}{2}$$ $\int_0^{2 \pi} \sin^2 $$\Theta$$ d \Theta \: + \: \int_0^{2 \pi} \sin^2 $$\Theta$$ d \Theta$
\;
y_{avg} \: = \: $$\frac{1}{2 \pi}$$ $$\frac{1}{2}$$ $\int_0^{2 \pi} \sin^2 $$\Theta$$ d \Theta \: + \: \int_0^{2 \pi} \cos^2 $$\Theta$$ d \Theta$
\;
y_{avg} \: = \: $$\frac{1}{2 \pi}$$ $$\frac{1}{2}$$ $\int_0^{2 \pi} $$\sin^2 \( \Theta$$ \: + \: \cos^2 $$\Theta$$ \) d \Theta$
\;
y_{avg} \: = \: $$\frac{1}{2 \pi}$$ $$\frac{1}{2}$$ \int_0^{2 \pi} 1 \: d \Theta
\;
y_{avg} \: = \: $$\frac{1}{2 \pi}$$ $$\frac{1}{2}$$ $$2 \pi$$
\;
y_{avg} \: = \: \frac{1}{2}
\)

#### MrAl

Joined Jun 17, 2014
10,620
Hi again,

Yes i meant to say calculate the average over one period which involves the integration. It is corrected now of course. Taking the average means integrating over the period and then dividing by the period. And yes, either sine or cosine.
This shows another way to calculate the average power.

#### e-learner

Joined Apr 25, 2015
30
Hello again,

I was throwing some ideas out there. I got the book result so i was suggesting some things that might have went wrong.

So what was your mistake again, and do you get the right result now?

It is also interesting that when you integrate sin(wt)^2 over one period and divided by the period you get 1/2.

As WBahn told that the voltage appears across the impedance and not the resisitor. I was calculating power taking that voltage across the resistive portion - this was my mistake.So, I wrote the impedance in rectangular form and calculated the voltage for the resistive portion using voltge divider and then calculated the average power

#### MrAl

Joined Jun 17, 2014
10,620
Hi again,

Oh that makes it clear now. Yeah, the source voltage can not appear across the resistor if there is another impedance in series with the resistor.