# Power in fullwave rectified circuit

1. Nov 2, 2009

### songoku

1. The problem statement, all variables and given/known data
A sinusoidal alternating current is fullwave rectified. The rectified current will produce in the same load

a. the same power
b. 0.71 times the power
c. 1.41 times the power
d. half the power
e. twice the power

2. Relevant equations
Pmax=Imax x Vmax

Paverage=Irms x Vrms

3. The attempt at a solution
I guess the question is asking about the average power. For full wave rectified :

$$I__rms =\frac{I_max}{\sqrt{2}}$$

$$V__rms =\frac{V_max}{\sqrt{2}}$$

Paverage=Irms x Vrms

= $$\frac{I_max}{\sqrt{2}}\times \frac{V_max}{\sqrt{2}}$$

= $$\frac{P_max}{2}$$

So, the answer is (d) ?

And how to determine whether the question is asking about max. or average power?

Thanks

2. Nov 2, 2009

### Delphi51

Go back to the basics. Power is the integral over a cycle of I*V or I²R dt.
Sketch the waveforms and shade in the area of the integral in each case.
How do the areas compare?

3. Nov 2, 2009

### songoku

Hi Delphi51
I tried to sketch the waveform of sine wave and full wave rectified, then compared the areas. The areas will be the same so the answer should be (a) ?

4. Nov 2, 2009

Yes.

5. Nov 2, 2009

### songoku

Hi Delphi51

Now I see that what I've done on the first post didn't answer the question. It's just a little work to show that the average power is half of the max. power.

Thanks a lot !!

6. Nov 2, 2009

### Delphi51

Yes, nothing to do with average power. Just a comparison between sine wave and rectified sine wave.