A three-phase star-connected synchronous alternator of rating 22 kVA, 20 kV, 50 Hz has synchronous reactance of 8 Ω per phase. The induced voltage per phase is 20 kV and the line terminal voltage is 15 kV. Find the 3-phase maximum power of the machine.

Correct option is B

$$\text{Given:} \\[4pt]\text{Line terminal voltage, } V_L = 15\,\text{kV} \\[4pt]\text{Synchronous reactance per phase, } X_s = 8\,\Omega \\[4pt]\text{Induced voltage per phase, } E_p = 20\,\text{kV} \\[8pt]\text{The 3-phase power in an alternator is given by:} \\[4pt]P = \frac{3 V_{ph} E_p \sin \delta}{X_s} \\[8pt]\text{For maximum power, the power angle } \delta = 90^\circ, \text{ thus } \sin \delta = 1. \\[4pt]\text{Therefore, the maximum power is:} \\[4pt]P_{\text{max}} = \frac{3 V_{ph} E_p}{X_s} \\[8pt]\text{Where:} \\[4pt]V_{ph} = \text{the phase voltage,} \\[4pt]E_p = \text{the induced voltage per phase,} \\[4pt]X_s = \text{the synchronous reactance.} \\[8pt]\textbf{Calculation:} \\[4pt]\text{To calculate the phase voltage } V_{ph}, \text{ use the line voltage relation for a star-connected system:} \\[4pt]V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{15}{\sqrt{3}} = 8.66\,\text{kV} \\[8pt]\text{Now, substitute the values into the formula for maximum power:} \\[4pt]P_{\text{max}} = \frac{3 \times 8.66 \times 20}{8} \\[4pt]P_{\text{max}} = 112.5\,\text{MW}$$​​