Transformer Maximum Efficiency Condition Explained

A transformer is a static electrical machine used to transfer power from one circuit to another without changing the frequency. The efficiency of a transformer is defined as the ratio of output power to input power. In an ideal transformer, efficiency would be 100%, but in real transformers, there are losses.

The input power to a transformer is equal to the output power plus the losses. Therefore, efficiency ($\eta$) can be expressed as:

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

The losses in a transformer consist of two main types:

• Iron Losses (Core Losses): These losses occur in the core of the transformer due to hysteresis and eddy currents. Iron losses ($P_i$) depend on the voltage and frequency and are practically constant regardless of the load.
• Copper Losses ($I^2R$ Losses): These losses occur in the windings due to the resistance of the copper wire. Copper losses ($P_{cu}$) are proportional to the square of the load current ($I_2^2$) and the winding resistance ($R_{eq}$ referred to the secondary). So, $P_{cu} = I_2^2 R_{eq}$. These losses vary with the load.

Let the secondary terminal voltage be $V_2$ and the load current be $I_2$. If the power factor of the load is $\cos\phi_2$, the output power is $V_2 I_2 \cos\phi_2$.

The total losses are $P_{\text{losses}} = P_i + P_{cu}$.

So, the efficiency can be written as:

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

To find the condition for maximum efficiency, we can differentiate the efficiency expression with respect to the variable load (represented by the load current $I_2$) and set the derivative to zero. Assuming $V_2$, $\cos\phi_2$, $P_i$, and $R_{eq}$ are constant for a specific operating condition (like full load voltage and a particular power factor), the efficiency varies only with $I_2$.

Maximum efficiency occurs when the variable losses (Copper Losses) equal the constant losses (Iron Losses).

Mathematically, maximum efficiency is achieved when:

$\frac{d\eta}{dI_2} = 0$

This differentiation yields the condition:

$I_2^2 R_{eq} = P_i$

Which means:

Copper Loss = Iron Loss

At maximum efficiency, the load current $I_{2(\text{max eff})}$ is such that $I_{2(\text{max eff})}^2 R_{eq} = P_i$.

Therefore, the transformer gives maximum efficiency when its copper losses are equal to its iron losses. This condition balances the constant core losses with the load-dependent copper losses to achieve the highest ratio of output to input power.

Let's look at the given options:

• When iron loss is greater than copper loss - This is not the condition for maximum efficiency.
• When iron loss is zero - Iron loss is a constant loss in a real transformer; it is not zero.
• When iron loss equals copper loss - As derived, this is the condition for maximum efficiency.
• When iron loss is half of the copper loss - This is not the condition for maximum efficiency.

Based on the analysis, the condition for a transformer to give maximum efficiency is when the iron loss is equal to the copper loss.