At light load transformer efficiency is less because

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Examoogle · Accepted Answer

fixed loss is high compared to output

Explanation: The efficiency of a transformer is defined as the ratio of output power to input power. At light loads, the constant core losses (hysteresis and eddy current losses) dominate the total power losses because the variable copper losses (which depend on the square of the current) are negligible.

Working Principle: Total losses in a transformer consist of iron losses ($P_i$) and copper losses ($P_{cu}$). Iron losses remain constant regardless of load, while copper losses vary with the square of the load current ($I^2R$). Efficiency $\eta = \frac{Output}{Output + P_i + P_{cu}}$. As the load decreases, the Output power decreases faster than the losses, causing the fixed core loss $P_i$ to become a significant percentage of the total energy, reducing overall efficiency.

Key Formulas: $\eta = \frac{V_2 I_2 \cos \phi_2}{V_2 I_2 \cos \phi_2 + P_i + I_2^2 R_{eq}}$ — General efficiency equation; $P_{cu} = I_2^2 R_{eq}$ — Copper loss which is load dependent

Key Points: Fixed (Iron) losses are independent of the load current.; Variable (Copper) losses are proportional to the square of the load current.; Maximum efficiency occurs when fixed losses equal variable losses ($P_i = P_{cu}$).; Light load operation results in a low efficiency ratio due to the fixed $P_i$ overhead.

Advantages: High efficiency at rated load; Low maintenance due to no moving parts

Disadvantages: Inefficient at very light loads; Requires proper load matching for optimal performance

Applications: Power distribution systems; Voltage level matching in transmission networks

Additional Info: Transformer efficiency is typically very high (95-99%) at full load.. Option D is incorrect because small copper loss is a consequence of light load, not the primary reason for low efficiency; the dominance of fixed losses is the primary cause.

Answer: C is correct — At light load, the constant core (iron) losses are significantly larger relative to the small output power, resulting in a poor efficiency ratio.

Exam Tip: Always remember: Maximum efficiency occurs when the variable copper loss exactly equals the constant iron loss. This is a common numerical problem in competitive exams.

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