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: At half load, copper loss \(= (0.5)^2 \times 800 = 200\) W; iron loss stays \(500\) W; total loss \(= 700\) W.
Step 2: Output power \(= 0.5 \times 40 \times 0.8 = 16\) kW \(= 16000\) W.
Step 3: Express losses as a fraction of the output alone (loss-to-output ratio): \( \dfrac{700}{16000} \approx 0.04375 \).
Step 4: Since input \(=\) output \(+\) losses, efficiency \(\approx \dfrac{1}{1+0.04375} \times 100\).
\[ \eta \approx \frac{1}{1.04375}\times 100 \approx \boxed{95.81\%} \]

Answer 2

A third way to check this is to build the answer from the input side rather than the output side: assume the transformer is delivering exactly its half-load output, work out what must be supplied to the primary to sustain that, and see which option matches.

Real output required at half load, 0.8 pf: \(16\) kW. Since losses must be supplied in addition to this output, the electrical input power is \(16 + 0.7 = 16.7\) kW, where the \(0.7\) kW total loss combines the fixed \(500\) W iron loss with the quarter-value \(200\) W copper loss at half current.

1. 92%: Would require an input of about \(16/0.92 \approx 17.39\) kW, implying losses of roughly \(1.39\) kW, well above the \(0.7\) kW this transformer actually dissipates at half load.
2. 90%: Would require an input of about \(17.78\) kW, implying an even larger loss of nearly \(1.78\) kW.
3. 89.31%: Would require an input of about \(17.92\) kW, implying losses of almost \(1.92\) kW, far more than the transformer's actual \(0.7\) kW loss at this load.
4. 95.81%: Requires an input of \(16/0.9581 \approx 16.70\) kW, implying losses of about \(0.70\) kW, matching the transformer's actual combined iron and copper loss at half load exactly.

Building the calculation from the input side and comparing implied losses against the transformer's actual \(700\) W loss again isolates the same option.

Therefore, the correct answer is 95.81%.