Correct Answer - Option 1 : 1.92 kV

__Concept:__

The induced emf in an alternator is given by the formula:

E = 4.44 × ϕ × f × N_{ph} × k_{c} × k_{d}

\({k_c} = \cos \frac{{n\alpha }}{2}\)

\({k_d} = \frac{{\sin \left( {\frac{{nm\beta }}{2}} \right)}}{{m\sin \left( {\frac{{n\beta }}{2}} \right)}}\)

ϕ is the flux per pole

f is the frequency

N_{ph} – turns per phase

k_{c} – chording factor

k_{d} – 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 (N_{s}) = 1200

Number of poles (P) = 6

\({N_s} = 1200 = \frac{{120 \times f}}{6}\)

⇒ f = 60 Hz

Turns per phase \(\frac{{54}}{{3 \times 2}} \times 8 = 72\)

K_{c} = K_{d} = 1

The no-load phase voltage in the alternator is,

E = 4.44 × 0.1 × 60 × 72 × 1 × 1

= 1.92 kV