If a transformer operating most efficiently at 2/3rd of full load then the ratio of full load
copper loss and iron loss is?

To determine the ratio of full load copper loss to iron loss for a transformer operating most efficiently at two-thirds of its full load, we need to understand the concepts of copper loss and iron loss in transformers. Let's break this down step by step.

Understanding Transformer Losses

Transformers experience two main types of losses: copper losses and iron losses.

• Copper Losses: These occur due to the resistance in the windings of the transformer and are proportional to the square of the load current. Mathematically, copper loss (P_cu) can be expressed as:
• P_cu = I²R, where I is the load current and R is the resistance of the windings.
• Iron Losses: Also known as core losses, these are primarily due to hysteresis and eddy currents in the core material. Iron losses remain relatively constant regardless of the load and are typically represented as a fixed value (P_fe).

Efficiency and Load Conditions

The efficiency of a transformer is maximized when the total losses are minimized. At two-thirds of full load, the copper losses will be different compared to full load. Let's denote the full load current as I_FL. At full load, the copper loss can be expressed as:

P_cu,FL = I_FL²R

At two-thirds of full load, the current becomes:

I_2/3 = (2/3)I_FL

Thus, the copper loss at two-thirds load is:

P_cu,2/3 = (2/3 I_FL)²R = (4/9)I_FL²R

Calculating the Loss Ratio

Now, we can calculate the ratio of full load copper loss to iron loss. The full load copper loss is:

P_cu,FL = I_FL²R

The iron loss remains constant at full load and at two-thirds load, denoted as P_fe. Therefore, the ratio of full load copper loss to iron loss can be expressed as:

Ratio = P_cu,FL / P_fe = I_FL²R / P_fe

At the point of maximum efficiency, the copper losses at two-thirds load will equal the iron losses. Therefore, we can set:

P_cu,2/3 = P_fe

Substituting the values, we find:

(4/9)I_FL²R = P_fe

Final Ratio Derivation

From the above, we can derive the ratio of full load copper loss to iron loss:

Ratio = P_cu,FL / P_fe = (I_FL²R) / (4/9)I_FL²R = 9/4

Thus, the ratio of full load copper loss to iron loss is 9:4. This means that at full load, the copper losses are significantly higher than the iron losses, but at the point of maximum efficiency (two-thirds load), they balance out, leading to optimal performance.