# half-wave rectifier with delay angle

Discussion in 'Homework Help' started by notoriusjt2, Oct 3, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0

it cant be this easy... what formula should I be using?

2. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771

Delay angle or not. A resistor will dissipate power according to the applied RMS Voltage
Formula is P=V^2/R

3. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
in the book it gives me these equations....

Vm=Vpeak
then it says to use the P=Vrms^2/R

but it gives me no formulas that have anything to do with current.
are you sure thats still a valid answer?

4. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771
What does that equation look like.
Same one I gave u...forget the current. It is their to confuse newbies like you, and it is doing a pretty good job too.

5. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
haha i gotcha, it is very confusing. that Vrms calculation takes into account for the delay angle. alpha=delay angle

so the delay angle has nothing to do with it? it doesnt even matter?

6. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771
Delay angle matters when the load is Reactive, Like Capacitive or Inductive.

Besides, ur Q, says average current.
Average values are not used for power calculation. Didn't u know that?

7. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
i went with answer 'E' and it was incorrect

i know that delay angle needs to be accounted for somewhere

8. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771
shoooooot!!!. I missed the half wave rectifier part..
That was really clumsy of me...
Hold on, let me see

9. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771
Answer is way wrong. Since the AC is rectified we have to find the voltage seen by the load. which is lower than Vrms.
In the formula shows Vo, is any other formula given for Vo

{ed}
Is this q related to practical application. The delay angle is the SCR firing angle to rectify the AC. You have to find out Vo using a scope if this is practical based.

Last edited: Oct 3, 2010
10. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
You must first find the conduction interval that will give an average current of 2.5 amps into a 30Ω resistor.

Having found that conduction interval you then integrate instantaneous product of voltage and current over the conduction interval. That will give you the power absorbed by the resistor.

Alternatively, having found the relevant conduction interval, you could calculate the RMS value of that voltage coming out of the controlled rectifier, and use that value in the familiar formula P = E^2/R. The result is the same.

The first attached image shows the details. The second shows the RMS calculation mentioned in the paragraph above.

File size:
23.3 KB
Views:
97
File size:
5.7 KB
Views:
61
11. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,785
771
Can u get the right answer stated in the q?. electrician?

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
The correct answer would appear to be (c).