Question

A single phase transformer is rated at 40 kVA. The transformer has full-load copper losses of 800 W and iron losses of 500 W. Determine the transformer efficiency at half full-load and 0.8 power factor.

• A. 92%
• B. 90%
• C. 95.81%
• D. 89.31%

Hint:

Copper losses depend on load current, while iron losses remain constant. Efficiency calculations must account for both.

Answer 1

The Correct Option is C

Step 1: Determine the output power at half full-load.
Rated apparent power of the transformer is
\[ S = 40 \text{ kVA} \]
At half full-load,
\[ S_{\text{half-load}} = 0.5 \times 40 = 20 \text{ kVA} \]
Given power factor is 0.8, therefore output power is
\[ P_{\text{out}} = 20 \times 0.8 = 16 \text{ kW} \]
Step 2: Calculate iron losses.
Iron losses are constant and independent of load.
\[ P_{\text{iron}} = 500 \text{ W} \]
Step 3: Calculate copper losses at half full-load.
Copper losses vary as the square of the load.
\[ P_{\text{cu, half-load}} = (0.5)^2 \times 800 = 200 \text{ W} \]
Step 4: Calculate total losses at half full-load.
\[ P_{\text{loss}} = 500 + 200 = 700 \text{ W} \]
Step 5: Calculate transformer efficiency.
\[ \eta = \frac{P_{\text{out}}}{P_{\text{out}} + P_{\text{loss}}} \]
\[ \eta = \frac{16}{16 + 0.7} = 0.9581 \]
\[ \eta = 95.81% \]
Step 6: Conclusion.
The efficiency of the transformer at half full-load and 0.8 power factor is
\[ \boxed{95.81%} \]