Question

A single-phase transformer has maximum efficiency of 98%. The core losses are 80 W and the equivalent winding resistance as seen from the primary side is 0.5 \( \Omega \). The rated current on the primary side is 25 A. The percentage of the rated input current at which the maximum efficiency occurs is

• A. 35.7%
• B. 50.6%
• C. 80.5%
• D. 100%

Hint:

For maximum efficiency in a transformer, the copper losses must equal the core losses. Use this condition to find the operating current at maximum efficiency.

Answer 1

The Correct Option is B

For maximum efficiency, the condition is that the copper losses equal the core losses.

Step 1: Calculate copper losses.
The copper loss \( P_{\text{cu}} \) is given by: \[ P_{\text{cu}} = I_{\text{load}}^2 \times R, \] where \( I_{\text{load}} \) is the load current and \( R \) is the equivalent winding resistance.

Step 2: Set up the maximum efficiency condition.
At maximum efficiency, copper losses equal core losses, i.e.: \[ P_{\text{cu}} = P_{\text{core}} = 80 \, \text{W}. \] Substituting into the copper loss formula: \[ I_{\text{load}}^2 \times 0.5 = 80. \]

Step 3: Solve for \( I_{\text{load}} \).
Solving for \( I_{\text{load}} \): \[ I_{\text{load}}^2 = \frac{80}{0.5} = 160. \] \[ I_{\text{load}} = \sqrt{160} = 12.65 \, \text{A}. \]

Step 4: Calculate the percentage of rated current.
The rated current is given as 25 A. The percentage of the rated current is: \[ \text{Percentage} = \frac{I_{\text{load}}}{I_{\text{rated}}} \times 100 = \frac{12.65}{25} \times 100 = 50.6%. \] Thus, the percentage of the rated input current at which the maximum efficiency occurs is \( 50.6% \).

Final Answer: 50.6%