How to Calculate Average Power Dissipation on a Resistor?

In summary, the conversation discussed finding the average power dissipated on a 14 kΩ resistor when given a specific voltage signal. The equation Pavg = (sqrt(Vdc^2 + vp^2/2))/R was mentioned, but it was noted that this was incorrect and the correct method involved combining the two cos() terms to get a resultant AC waveform. The use of Phasors was suggested, but it was later discovered that the frequencies of the AC components were actually off by a factor of two. The correct method was then found to be using the power spectrum and dividing the individual signal powers by 2 (Vrms^2/2). The reason for this division was not known.
  • #1
eatsleep
42
0
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.
 
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  • #2
eatsleep said:
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.

What's vp? How did you combine the two cos() terms to get one AC term?
 
  • #3
berkeman said:
What's vp? How did you combine the two cos() terms to get one AC term?

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
eatsleep said:
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.

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
berkeman said:
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...

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
eatsleep said:
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?

Oh, oops, you are right. They are off by a factor of two.
 
  • #7
berkeman said:
Oh, oops, you are right. They are off by a factor of two.

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
 

What is average power and why is it important?

Average power is the amount of energy transferred or used per unit of time. It is important in various scientific fields, such as physics, engineering, and mathematics, as it helps to quantify the rate at which energy is being transferred or used.

How do you calculate average power?

The formula for calculating average power is P = W/t, where P is the average power, W is the work done, and t is the time taken. Alternatively, it can also be calculated by dividing the energy used by the time taken.

What are the units of average power?

The SI unit for average power is watts (W). However, other common units include horsepower (hp) and kilowatts (kW).

How is average power different from instantaneous power?

Average power is the average amount of energy used or transferred over a period of time, while instantaneous power is the amount of energy being used or transferred at a specific moment in time. Average power takes into account the entire time period, while instantaneous power only considers a single point in time.

What factors can affect the calculation of average power?

The calculation of average power can be affected by various factors such as the efficiency of the system, the type of energy being used, and any external factors that may affect the rate of energy transfer or usage. It is important to consider all these factors in order to accurately calculate average power.

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