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प्रश्न
An ac generator generates an emf which is given by e = 311 sin (240 πt) V. Calculate:
- frequency of the emf.
- r.m.s. value of the emf.
उत्तर
(1) Given that
e = 311 sin (240 πt) V ...(i)
∵ e = e0 sin (ωt) V ...(ii)
From equations (i) and (ii)
e0 = 311 and ωt = 240 πt
∴ ω = 240 π
f = `omega/(2pi)`
`= (240 pi)/(2pi)`
= 120 cycle/s
e0 = Peak emf of the generator = 33
∵ Peak emf = rms · emf × `sqrt2`
∴ rms = `1/sqrt2` peak emf
= 0.707 × 311 = 219.87 volts
∴ rms = 219.87 volts
(2) Given the number of turns in the primary winding is NP = 60 turns.
The number of turns in secondary winding is NS = 3000 turns.
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