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.
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).
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).
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