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?
Discussion:
16 comments Page 2 of 2.
Sai Kumar said:
1 decade ago
Half wave rectifier.
Vdc = Vpeak/pi.
For full wave rectifier.
Vdc = 2*Vdc (half wave rectifier).
54*2 = 108.
Vdc = Vpeak/pi.
For full wave rectifier.
Vdc = 2*Vdc (half wave rectifier).
54*2 = 108.
(1)
Gokul said:
1 decade ago
In the question they ask for two full wave's average peak value. So it will be 54*4=216. Am I correct?
Shadan said:
8 years ago
Vm/π :- Half wave avvg voltage.
Full wave avrg, voltage=2Vm/π.
So 2*169.7/π
= 108.
Full wave avrg, voltage=2Vm/π.
So 2*169.7/π
= 108.
(1)
Nirmala said:
7 years ago
54 = vm/π
vm = 54*π
fwr = 2*vm/π
= 2 * 54*π/π = 2 * 54 = 108v.
vm = 54*π
fwr = 2*vm/π
= 2 * 54*π/π = 2 * 54 = 108v.
(3)
M.lakshmireddy said:
5 years ago
169.7(54).
399.4(?),
?=54*399.4/169.7 = 108.0v.
399.4(?),
?=54*399.4/169.7 = 108.0v.
Prince said:
4 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.
So, the solution becomes 2(54) = 108.0V.
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