Power of voltage series

[BamaBino]

The answer is suppose to be B.

Would someone give the correct solution?

Is this type problem fair-game for the Power exam?

Thanks.

 

[Adrenaline]

Your error is being made by the fact you cannot apply superposition to multiple sine wave sources of the same frequency but different phases.

If you plot 4sqrt(2)cos(120*pi*t)-6cos(120*pi*t+30) you will see it has little power.

I have the pertinent equation at home I can get at lunch time.

Maybe someone else will provide it in the mean time.

Edit: Actually you can just do it with phasor notation.

First note the power from the 3 volt source and the 2cos(240*pi*t) source is 5mW.

The rest is from the cos(120*pi*t) sources

(5.65 - 6<30) = 3.034<-81

3*3/2/2200 = 2.04mW

So...

(3^2+(2^2+(3.035)^2)/2)/2200 = 7.045mW

 

[mudpuppy]

I certainly can't guarantee anything, but I wouldn't expect a problem like this on the Power exam. It looks more like a communications question to me. I expect questions on the power exam to be more in the MW range than the mW range.

 

[benbo]

I certainly can't guarantee anything, but I wouldn't expect a problem like this on the Power exam. It looks more like a communications question to me. I expect questions on the power exam to be more in the MW range than the mW range.

I agree. In fact that looks sort of like a problem I recognize from studying for that communications exam many moons ago.

 

[Adrenaline]

I did not see anything similar on the Electronics exam. However, there was one in the practice test from NCEES, so I think it is fair game at least for the Electronics exam.