Electronics - Alternating Current and Voltage - Discussion
Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 6)
6.
What is the instantaneous peak voltage at 250° on a 6 V peak sine wave?
Discussion:
11 comments Page 1 of 2.
Ancy said:
5 years ago
Because θ =wt.
Ajit said:
5 years ago
How wt is equal to 250? Please explain.
Sam said:
9 years ago
Formula = Io sin2^ft.
F = 250/2.
= 6sin 250,
= 5.638 = -5.64,
F = 250/2.
= 6sin 250,
= 5.638 = -5.64,
(1)
VamsI said:
9 years ago
Excellent explanation. Thank you all.
Ajinkya said:
9 years ago
Thank you, all for your answer and explanation.
PrivUsr said:
10 years ago
The instantaneous values of a sinusoidal waveform is given as:
Vi = Vmax Sin θ.
= 6 Sin 250°
= 6(-.94).
= -5.64 V.
Vi = Vmax Sin θ.
= 6 Sin 250°
= 6(-.94).
= -5.64 V.
Arya said:
1 decade ago
Please give the answer.
Subash Pakhrin said:
1 decade ago
V = Vmax*Sin(wt).
Where wt is angular frequency in this case is 250 so,
V = 6 * sin(250).
Therefore V = -5.64.
Where wt is angular frequency in this case is 250 so,
V = 6 * sin(250).
Therefore V = -5.64.
Yash said:
1 decade ago
According to the formula, let
V [instantaneous] = V[peak]sin[wt].
6*sin[256].
6*-0.097.
-5.64.
V [instantaneous] = V[peak]sin[wt].
6*sin[256].
6*-0.097.
-5.64.
Ram said:
1 decade ago
V=Vm Sin wt so, V=6,wt=250
V=6 sin250
V=-5.64v
V=6 sin250
V=-5.64v
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