Answer :

244.5 (Range : 240 to 248)

Concept: The excitation in synchronous motor for a leading phase is given by the expression

$E=\sqrt{(Vcos\varphi -I_a{R}_a{)}^2+(Vsin\varphi +I_a{X}_s{)}^2}$............(1)

Also, the power is given as

$P_{in}=\sqrt{3}{V}_L{I}_{L}cos\varphi$........(2)

Input = Output-losses ..........(3)

Calculation:

Given:

losses=2 kW

Output=10 kW

⇒ Input=10 + 2 = 12 kW

Using eqn (2),

$I_L=\frac{12000}{\sqrt{3}\times 400\times 0.8}=21.65A=I_a$

Put the above-calculated current with other given values of p.f, resistances, and reactance in Eqn (1)

$I_L=\frac{12000}{\sqrt{3}\times 400\times 0.8}=21.65A=I_a$

$E=\sqrt{((\frac{400}{\sqrt{3}})\times 0.8-21.65\times 0{)}^2+(\frac{400}{\sqrt{3}}\times 0.6+21.65\times 1{)}^2}$

⇒ E=244.55V