Question

A transformer has full load copper loss of 400 W. The copper loss at half full load will be

• A. \( 400\text{ W} \)
• B. \( 200\text{ W} \)
• C. \( 100\text{ W} \)
• D. \( 50\text{ W} \)

Hint:

Copper loss scales quadratically with the load factor ($x^2$). - At $\frac{1}{2}$ load $\Rightarrow$ loss becomes $\frac{1}{4}$ of full load. - At $\frac{1}{3}$ load $\Rightarrow$ loss becomes $\frac{1}{9}$ of full load. - At $2$ times load $\Rightarrow$ loss becomes $4$ times full load.

Answer 1

The Correct Option is C

Concept: The total losses in a transformer are broadly divided into two main categories: core losses (iron losses) and copper losses ($I^2R$ losses).
• Iron Losses ($P_i$): These occur in the magnetic core due to hysteresis and eddy currents. They depend on the supply voltage and frequency and remain constant regardless of the electrical load connected to the secondary winding.
• Copper Losses ($P_{cu}$): These occur due to the ohmic resistance of the primary and secondary copper windings. Since power dissipated in a resistor is proportional to the square of the current ($P = I^2R$), copper losses vary directly with the square of the load current. If the load scales by a fraction $x$ (where $x = \frac{\text{Actual Load}}{\text{Full Load}}$), the current scales by $x$, and consequently, the new copper loss can be computed as: $$P_{cu,\text{ new}} = x^2 \cdot P_{cu,\text{ full-load}}$$

Step 1: Identify the given values and parameters.
We are given the full-load copper loss value: $$P_{cu,\text{ full-load}} = 400\text{ W}$$ We need to find the copper loss at "half full load". This implies that the loading factor $x$ is: $$x = \frac{1}{2} = 0.5$$

Step 2: Apply the mathematical scaling law for copper losses.
Since copper loss is dependent on the square of the current, let us express it in terms of the load factor $x$: $$P_{cu}(x) = x^2 \cdot P_{cu,\text{ full-load}}$$ Substituting $x = \frac{1}{2}$ into the formula: $$P_{cu}\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 \cdot 400$$

Step 3: Perform the final numerical computation.
Squaring the fraction yields: $$\left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ Now, multiply this fraction by the full-load value: $$P_{cu}\left(\frac{1}{2}\right) = \frac{1}{4} \cdot 400 = 100\text{ W}$$ Thus, at half the full load, the copper loss drops significantly down to $100\text{ W}$. This perfectly corresponds to option (3).