Eng Pro Guides Final Exam #45 - Full Wave Rectifier

[DLD PE]

I solved this problem a different way.  For some reason I don't get the same solution if I follow the author's method:

The problem states, "A single phase, full wave rectifier has an input of 480V RMS.  What is the DC voltage to the load, assuming a 10 degree firing relay?

Author's solution:

Finding the output of the DC load, Vmax=480*sq.rt 2 = 678.823 (the author gets 678.72).

Then to get the DC voltage load, Vdc, integrate the sine function from 10 to 180 degrees.

Vdc = 1/pi x integral from 10 to 180:  687*sin(theta)V = 428.  

I don't understand why he's putting the Vmax amount inside the integral function.  If I do that, I get a crazy huge number, like 24,542V

My solution:

The average DC voltage is:  Vavg=  [480*sq.rt 2 / 180 degrees] x [ integral from 10 to 180 sin(x)dx] and I get 428.9.

Am I doing this wrong?  I'm getting the correct answer but trying to understand if there is a different way.  Just curious if anyone else has had trouble with it.

 

[BSNR]

Lookup the constant multiple rule for definite integrals. The constant can be pulled to the outside of the integral or left inside. The answer will be the same. 

Also, I just noticed the solution is mixing radians and degrees. The pi in the solution should be changed to 180 degrees or the degrees should be changed to radians. Just make sure your calculator is in the correct mode: degrees or radians.