A single-phase full-wave bridge rectifier has a resistive load of 18ohms and an ac source of 120V rms. Determine the average power in the load.

these average power questions are killing me....

Vo=((2)(120)(1.414))/pi = 108v
Io= Vm/R = 108/18 = 6A

P=108*6=648w

so what step am i forgetting to do here... i am never taking an electronics/math class online again.

 

I think that you need to find RMS voltage and current.

 

so
Vrms=108/1.414=76.37
Irms=6/1.414=4.24

P=76.37*4.24=324w

would this be correct?

 

\(P_{avg}=\left\langle \frac{V^2}{R} \right\rangle \\
=\frac{\langle V^2 \rangle}{R}\\
=\frac{V_{rms}^2}{R} \\
where\ \langle\ \rangle\ \text{is the time mean}\)

Your answer is wrong, because Vrms is not connected to the average voltage with the relation you wrote. The Vrms of the rectified waveform is the same as the Vrms of the initial sine wave. Do the math and it will become apparent. Use the default RMS calculation formula.

Is that clear?

 

Hehe, No.
Maybe you should read the question again.
"A single-phase full-wave bridge rectifier has a resistive load of 18ohms and an ac source of 120V rms. Determine the average power in the load."

 

\(P_{avg}=\left\langle \frac{V^2}{R} \right\rangle \\
=\frac{\langle V^2 \rangle}{R}\\
=\frac{V_{rms}^2}{R} \\
where\ \langle\ \rangle\ \text{is the time mean}\)

Your answer is wrong, because Vrms is not connected to the average voltage with the relation you wrote. The Vrms of the rectified waveform is the same as the Vrms of the initial sine wave. Do the math and it will become apparent. Use the default RMS calculation formula.

Is that clear?

yea that helps, but what do you mean when you say time mean?

so if Pavg=(V^2rms)/R
then
Pavg=(120^2)/18
Pavg=800w

is that what you mean?

 

Time mean is just the mean of a value over a time period, mean for short. Don't bother about it.

Other than that, your answer seems correct.