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.
Prashant said:
1 decade ago
@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.
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.
Sabelo said:
1 decade ago
For half wave: Vavg=Vp/pi.
54v = 169, 7/3, 14.
For full wave: Vavg=2Vp/pi.
= 2(169,7)/3, 14 = 108.
54v = 169, 7/3, 14.
For full wave: Vavg=2Vp/pi.
= 2(169,7)/3, 14 = 108.
Dinesh said:
1 decade ago
For half wave is = 169.7 and average voltage = 54.
= (54*2=108).
= (54*2=108).
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?
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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