Question

An alternator having an induced emf of 1.6 p.u. is connected to an infinite bus of 1 p.u. If the busbar has reactance of 0.6 p.u. and alternator has reactance of 0.2 p.u., what is the maximum power that can be transferred?

• A. 2 p.u.
• B. 2.67 p.u.
• C. 5 p.u.
• D. 6 p.u.

Hint:

The maximum power transfer in a synchronous machine can be calculated using the formula \( P_{\text{max}} = \frac{E_{\text{gen}} \cdot E_{\text{bus}}}{X_{\text{gen}} + X_{\text{bus}}} \), where \( E_{\text{gen}} \) is the generated emf and \( X_{\text{gen}}, X_{\text{bus}} \) are the reactances.

Answer 1

The Correct Option is A

The maximum power that can be transferred in a synchronous generator is given by the formula: \[ P_{\text{max}} = \frac{E_{\text{gen}} \cdot E_{\text{bus}}}{X_{\text{gen}} + X_{\text{bus}}} \] Where: - \(E_{\text{gen}} = 1.6 \, \text{p.u.}\) (Induced emf),
- \(E_{\text{bus}} = 1 \, \text{p.u.}\) (Infinite bus voltage),
- \(X_{\text{gen}} = 0.2 \, \text{p.u.}\) (Alternator reactance),
- \(X_{\text{bus}} = 0.6 \, \text{p.u.}\) (Busbar reactance).
Substituting the values: \[ P_{\text{max}} = \frac{1.6 \times 1}{0.2 + 0.6} = \frac{1.6}{0.8} = 2 \, \text{p.u.} \] Thus, the maximum power that can be transferred is 2 p.u., which corresponds to option (1).