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प्रश्न
An alternating voltage is given by e = 8sin628.4t. Find
- peak value of e.m.f.
- frequency or e.m.f.
- instantaneous value of e.m.f. at time t = 10. ms
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उत्तर
Given:
e = 8 sin (628.4 t) ...(i)
To find:
- peak value of e.m.f.
- frequency of e.m.f.
- instantaneous value of e.m.f. at time t = 10 ms.
We know, e = e0 sin(ωt) ...(ii)
Comparing equations (i) and (ii),
e0 = 8 V and ω = 628.4
1. Peak value of e.m.f.
e0 = 8 V
2. Frequency of e.m.f.
`f = omega/(2 pi) = 628.4/(2 xx 3.142) = 100 Hz`
3. Instantaneous value of e.m.f. at time
t = 10 ms = 10 × 10−3s
Substituting in equation (i),
e = 8 sin [(628.4) × 10 × 10−3]
= 8 sin (6284 x 10−3)
= 8 sin (6.284)
= 8 sin (2 π)
∴ e = 0 V
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