Electrical Engineering - Alternating Current and Voltage - Discussion
Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 22)
22.
A waveform has a baseline of 3 V, a duty cycle of 20%, and an amplitude of 8 V. The average voltage value is
Discussion:
20 comments Page 2 of 2.
ROOPESH said:
1 decade ago
Integrate 8 dT from 0 to 0.2 and integrate 3 dT from 0.2 to 1. Then add the two results.
So we get (8*0.2)+(3*0.8) = 4.
So we get (8*0.2)+(3*0.8) = 4.
Amir said:
1 decade ago
Any one explain baseline kindly?
PINKY said:
1 decade ago
Please explain what is base line?
Gopal said:
1 decade ago
I am not yet understand the meaning of base line, please anyone explain.
Mohan said:
1 decade ago
@Giri.
The baseline value may be back EMF value. So we should consider the waveform from above baseline value.
The baseline value may be back EMF value. So we should consider the waveform from above baseline value.
Sivanath said:
1 decade ago
Answer would be B.
4.6 v because I have a reason 3 v be dc wave and (baseline +(p-p*duty cycle)) = (3+(8*0.2)) = 3+1.6 = 4.6 v.
4.6 v because I have a reason 3 v be dc wave and (baseline +(p-p*duty cycle)) = (3+(8*0.2)) = 3+1.6 = 4.6 v.
Giri said:
1 decade ago
Can any one tell me what is meant by baseline voltage?
Mrakovic said:
1 decade ago
Baseline + (duty cycle*p-p amplitude) = Average voltage.
Baseline = 3V.
Duty cycle = 0.2.
p-p amplitude = amplitude-baseline = 8V-3V = 5V.
3 + (0.2*5) = 3+1 = 4V average voltage.
Baseline = 3V.
Duty cycle = 0.2.
p-p amplitude = amplitude-baseline = 8V-3V = 5V.
3 + (0.2*5) = 3+1 = 4V average voltage.
(1)
Chandrasekar said:
1 decade ago
@Drishti: why do u integrate 3dT from 0 to .8T ?? you can simply add 3 with (8*0.2), i guess the answer is 4.6
Drishti said:
1 decade ago
Integrate 8dT from 0 to 0.2T with period T and integrate 3dT from 0 to 0.8T with period T. Then add the two results.
We get 8*0.2 + 3*0.8 = 4.
We get 8*0.2 + 3*0.8 = 4.
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