Transformer Efficiency Explained

Understanding the efficiency of a single-phase transformer is crucial for electrical engineering applications. Efficiency ($\eta$) measures how effectively a transformer converts electrical power from the primary side to the secondary side. It is defined as the ratio of output power to input power.

The formula for efficiency is:

$$ \eta = \frac{\text{Output Power}}{\text{Input Power}} = \frac{\text{Output Power}}{\text{Output Power} + \text{Losses}} $$

Transformer losses are primarily categorized into two types:

• Core Losses (or Iron Losses): These are constant losses that occur in the transformer core. They include hysteresis losses and eddy current losses. These losses depend on the voltage and frequency but are independent of the load current. They are often denoted as P_core.
• Copper Losses (or Variable Losses): These losses occur in the transformer windings due to the resistance of the copper windings and the current flowing through them. They are proportional to the square of the load current ($I^2R$). They are denoted as P_cu.

Condition for Maximum Efficiency

The efficiency of a transformer is maximum when the variable losses (Copper losses) are equal to the constant losses (Core losses).

Mathematically, maximum efficiency occurs when:

$$ P_{cu} = P_{core} $$

This condition determines the load level (as a fraction of full load) at which the transformer operates most efficiently. Let $k$ be the load fraction, then $P_{cu} = k^2 \times P_{cu(FL)}$, where $P_{cu(FL)}$ is the full load copper loss. Thus, maximum efficiency occurs at $k = \sqrt{\frac{P_{core}}{P_{cu(FL)}}}$.

Efficiency at Constant Load Current and Power Factor

The question specifically asks about the condition for maximum efficiency when the load current is constant. Let's analyze the efficiency formula under this condition:

$$ \eta = \frac{V_2 I_2 \cos \phi_2}{V_2 I_2 \cos \phi_2 + P_{core} + I_2^2 R_{eq}} $$

Here:

• V₂ is the secondary voltage.
• I₂ is the secondary current.
• cos φ₂ is the load power factor.
• P_core is the constant core loss.
• I₂² R_eq is the copper loss, which is also constant because the load current ($I_2$) is stated to be constant.

In this specific scenario where both the load current ($I_2$) and the applied voltage (which determines $P_{core}$) are constant, the denominator of the efficiency formula ($P_{in} = P_{out} + P_{core} + P_{cu}$) has constant loss terms ($P_{core}$ and $I_2^2 R_{eq}$). The output power is given by $P_{out} = V_2 I_2 \cos \phi_2$.

To maximize the efficiency ($\eta = \frac{P_{out}}{P_{in}}$), given that the losses ($P_{core} + I_2^2 R_{eq}$) are constant, we need to maximize the output power ($P_{out}$). Since $V_2$ and $I_2$ are constant, the output power is directly proportional to the power factor ($\cos \phi_2$).

Therefore, the output power is maximized when the power factor ($\cos \phi_2$) is maximum.

The maximum value for the power factor is 1 (unity power factor).

Conclusion

When a single-phase transformer operates with a constant load current, its efficiency is highest when the power factor is at its maximum value, which is unity.