Question

A 5000 kVA, 1100 V, 50 Hz, Y-connected 3-phase alternator has armature resistance of 0.1 $\Omega$/phase and synchronous reactance of 1.5 $\Omega$/phase. Find the generated emf per phase, power factor is unity.

• A. 769.2 V
• B. 832.6 V
• C. 692.4 V
• D. 935.3 V

Hint:

In alternators operating at unity power factor, the generated emf is obtained by phasor addition of terminal voltage and synchronous impedance drops.

Answer 1

The Correct Option is B

Step 1: Calculate the rated line current.
\[ I_L = \frac{S}{\sqrt{3} V_L} \]
\[ I_L = \frac{5000 \times 10^3}{\sqrt{3} \times 1100} = 2624 \text{ A} \]
Step 2: Determine phase voltage and phase current.
For star connection,
\[ V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{1100}{\sqrt{3}} = 635.1 \text{ V} \]
\[ I_{ph} = I_L = 2624 \text{ A} \]
Step 3: Calculate voltage drops.
\[ I R_a = 2624 \times 0.1 = 262.4 \text{ V} \]
\[ I X_s = 2624 \times 1.5 = 3936 \text{ V} \]
Step 4: Calculate generated emf per phase.
At unity power factor,
\[ E_{ph} = \sqrt{(V_{ph} + I R_a)^2 + (I X_s)^2} \]
\[ E_{ph} = \sqrt{(635.1 + 262.4)^2 + (393.6)^2} \]
\[ E_{ph} = \sqrt{(897.5)^2 + (393.6)^2} \]
\[ E_{ph} = 832.6 \text{ V} \]
Step 5: Conclusion.
The generated emf per phase is
\[ \boxed{832.6 \text{ V}} \]