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 1 of 2.
Vipul chaudhary said:
1 decade ago
@Subbu is right,
Because .
Half wave peak=169.7 V, then full wave peak = 169.7*2 = 339.4 V.
and of two full wave peaks=2*339.4 =678.8
now,
average of Half wave peak=169.7 V is 54 of
and two(2)full wave peak=678.8/169.7 =4
therefore 4*54=216
therefor average of two full wave= 216/2=108
that means 1 full wave has 108
clue:in question there are two full wave given
Because .
Half wave peak=169.7 V, then full wave peak = 169.7*2 = 339.4 V.
and of two full wave peaks=2*339.4 =678.8
now,
average of Half wave peak=169.7 V is 54 of
and two(2)full wave peak=678.8/169.7 =4
therefore 4*54=216
therefor average of two full wave= 216/2=108
that means 1 full wave has 108
clue:in question there are two full wave given
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.
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).
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).
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.
Komal said:
1 decade ago
Half wave peak=169.7 V, then full wave peak = 169.7*2 = 339.4 V.
If 169.7 V has avg voltage=54 V,
Then, 339.4 V has avg volt = (54/169.7)*339.4 = 108 V.
If 169.7 V has avg voltage=54 V,
Then, 339.4 V has avg volt = (54/169.7)*339.4 = 108 V.
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?
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.
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)
Fathima said:
1 decade ago
I have also doubt that in question asked that what is the average of two full-wave peaks?
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