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.
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)
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)
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)
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.
Rik said:
1 decade ago
I have a bit doubt in this question can someone elaborate it please.
Fathima said:
1 decade ago
I have also doubt that in question asked that what is the average of two full-wave peaks?
Arvind kumar said:
1 decade ago
For half wave 54.0 So full wave 54*2=108
SUBBU said:
1 decade ago
Generally Full Wave Having 2 Half Waves. So, 54*2=108.
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
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).
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers