Correct Answer - Option 1 : 1.92 kV
Concept:
The induced emf in an alternator is given by the formula:
E = 4.44 × ϕ × f × Nph × kc × kd
\({k_c} = \cos \frac{{n\alpha }}{2}\)
\({k_d} = \frac{{\sin \left( {\frac{{nm\beta }}{2}} \right)}}{{m\sin \left( {\frac{{n\beta }}{2}} \right)}}\)
ϕ is the flux per pole
f is the frequency
Nph – turns per phase
kc – chording factor
kd – distribution factor
β – slot angle (electrical)
m – slots per pole per phase
α – chording angle
Calculation:
Given that, flux (ϕ) = 0.1 Wb
Slots = 54
Turns in each coil = 8
Speed (Ns) = 1200
Number of poles (P) = 6
\({N_s} = 1200 = \frac{{120 \times f}}{6}\)
⇒ f = 60 Hz
Turns per phase \(\frac{{54}}{{3 \times 2}} \times 8 = 72\)
Kc = Kd = 1
The no-load phase voltage in the alternator is,
E = 4.44 × 0.1 × 60 × 72 × 1 × 1
= 1.92 kV