# Calculating Average Power

1. Mar 18, 2013

### eatsleep

1. Given that v(t) = -6 + 12 cos(2π200t + π/4) - 9 cos(2π400t + π/6) volts. The signal is applied to a 14 kΩ resistor. Find the average power dissipated on the resistor in mW. Round your answer off to two decimal places.

2. Pavg = (sqrt(Vdc^2 + vp^2/2))/R

3. I used the above equation just P=Vrms^2/R to calculate the power, this is wrong and I'm not sure what to do.

Last edited by a moderator: Mar 18, 2013
2. Mar 18, 2013

### Staff: Mentor

What's vp? How did you combine the two cos() terms to get one AC term?

3. Mar 18, 2013

### eatsleep

I did not combine them i just added them together. vp is meant to be the the voltage peak. It would be 12 and -9.

4. Mar 18, 2013

### Staff: Mentor

That's not correct. You need to combine the AC components to get a resultant AC waveform. What if the phases of the two AC waveforms are exactly 180 degrees apart? You will not get a very big AC waveform then.

The frequencies of the AC components are the same, so you can use Phasors to solve this question...

5. Mar 18, 2013

### eatsleep

Is that the only way to do this question, I do not believe we have done phasors in class, also how are the frequencies the same?

6. Mar 18, 2013

### Staff: Mentor

Oh, oops, you are right. They are off by a factor of two.

7. Mar 18, 2013

### eatsleep

I just got the right answer I used power spectrum and did Vrms^2 and added all the individual signal powers up. On the sinusoidal waveforms I divided by 2 Vrms^2/2 I am not sure why, it said that in my notes. Why did they divide by 2?

Thanks for the help