Is the solution for this question correct?

[onemilimeter]

Pls advise whether the solution for the question shown in the attachment is correct or not. Thanks.

 

[ysenthilece]

ya the answer is correct ...
but two things look weird in the solution ...
1. cos is written as kos ..
2. cos is assumed and sin is drawn..

besides these ..the solution is perfect

 

[claudiocamera]

I have a different view, the answer is incorrect.

It was computed as if there was no a half wave rectifier.

To solve this question you don't need to going on complicated expressions. In a half wave rectifier Vdc = Vp/pi.
P=Vdc^2/R thus you have Vdc^2=30*24 =720.

Vdc=26.83 and Vp = 84.29.

As the frequency is 50 Hz you have: 84.29 *sin(100*pi*t)

 

[jallem]

I have a different view, the answer is incorrect.

No exactly, since the integration is done from 0 to Π, the answer
is correct. For a full sine Vrms=Vpk*0.707 and for a half wave
rectified sine Vrms = Vpk*0.5. (If I am not mistaken)
So, I think the answer is correct.

 

[onemilimeter]

Sorry... the 'kos' is 'cosine'.

Can we start the solution as follows:

V(t) = Vm Sin wt

 

[claudiocamera]

I have a different view, the answer is incorrect.

No exactly, since the integration is done from 0 to Π, the answer
is correct. For a full sine Vrms=Vpk*0.707 and for a half wave
rectified sine Vrms = Vpk*0.5. (If I am not mistaken)
So, I think the answer is correct.

Ok , I considered Vdc instead of Vrms , The answer is correct.