# Electronics - Diodes and Applications - Discussion

Discussion Forum : Diodes and Applications - General Questions (Q.No. 8)
8.
If a 169.7 V half-wave peak has an average voltage of 54 V, what is the average of two full-wave peaks?
119.9 V
108.0 V
115.7 V
339.4 V
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

Prince said:   3 years ago
Since the PIV of a half-wave is Vm and is the same 169.7V so finding the PIV of a full-wave will be 2Vm where Vm is 54v.

So, the solution becomes 2(54) = 108.0V.

M.lakshmireddy said:   4 years ago
169.7(54).
399.4(?),
?=54*399.4/169.7 = 108.0v.

Nirmala said:   6 years ago
54 = vm/π
vm = 54*π
fwr = 2*vm/π
= 2 * 54*π/π = 2 * 54 = 108v.
(3)

Vm/π :- Half wave avvg voltage.

Full wave avrg, voltage=2Vm/π.
So 2*169.7/π
= 108.
(1)

Gokul said:   9 years ago
In the question they ask for two full wave's average peak value. So it will be 54*4=216. Am I correct?

Sai Kumar said:   9 years ago
Half wave rectifier.
Vdc = Vpeak/pi.

For full wave rectifier.
Vdc = 2*Vdc (half wave rectifier).
54*2 = 108.
(1)

Dinesh said:   9 years ago
For half wave is = 169.7 and average voltage = 54.

= (54*2=108).

Sabelo said:   9 years ago
For half wave: Vavg=Vp/pi.

54v = 169, 7/3, 14.

For full wave: Vavg=2Vp/pi.

= 2(169,7)/3, 14 = 108.

@Jagdeep is right.

Vdc = Vm/pi for Half wave rectifier.

So Vdc = 169.7/ 3.14 = 0.318* 169.7 = 53.914 =~ 54.

And Vdc= 2Vm/pi for full wave rectifier So,

Vdc= (2*169.7)/ 3.14 = 107.9292 i.e. 108v.

Jagdeep Arry said:   1 decade ago
Here half wave rectification is:

Vdc = 0.318*V(peak).

e.g: 169.7*0.318 = 53.914 = 54(approx).

And for full wave rectification:

Vdc = 0.636*V(peak).

So,
Vdc = 0.636*169.7 = 107.9292 = 108V(approx).