Question

The copper and iron losses of a 1-$\phi$ transformer are equal at 90% of full load and its value is 162 W. The total loss at 80% of full load is

• A. 145 W
• B. 244 W
• C. 128 W
• D. 153 W

Hint:

To streamline this calculation without computing the full-load values explicitly, you can set up a direct ratio for the variable copper loss component: \[ P_{cu}(x_2) = P_{cu}(x_1) \cdot \left(\frac{x_2}{x_1}\right)^2 = 81 \cdot \left(\frac{0.8}{0.9}\right)^2 = 81 \cdot \frac{64}{81} = 64 \text{ W} \] Adding the constant iron loss of 81 W gives the total loss of 145 W quickly and efficiently!

Answer 1

The Correct Option is A

Concept: The total power loss ($P_{\text{total}}$) occurring inside an operating electrical transformer consists of two primary components:
• Iron Loss Core Loss ($P_i$): This includes hysteresis and eddy current losses within the magnetic core laminate. Because the operating flux density remains virtually constant under a fixed primary supply voltage and frequency, the iron loss is independent of the connected load current. It remains constant at all loading conditions ($P_i = \text{constant}$).
• Copper Loss Ohmic Loss ($P_{cu}$): This represents the $I^2R$ heating loss across the primary and secondary winding resistances. Because the current scales linearly with the fractional load $x$, the copper loss varies as the square of the fractional loading factor $x$: \[ P_{cu}(x) = x^2 \cdot P_{cu,\text{FL}} \] Where $x = \frac{\text{Actual Load}}{\text{Full Load}}$ and $P_{cu,\text{FL}}$ is the copper loss at full load condition ($x=1$). Let us solve the problem using these relationships over the two different loading stages:

Step 1: Extract parameters from the 90% load condition ($x_1 = 0.9$). The problem states that at 90% of full load ($x_1 = 0.9$), the copper loss and iron loss are equal, and each is valued at 162 W:
• Iron Loss: $P_i = 162 \text{ W}$
• Copper Loss at 90% load: $P_{cu}(0.9) = 162 \text{ W}$ Let us use this to find the full-load copper loss ($P_{cu,\text{FL}}$): \[ P_{cu}(0.9) = (0.9)^2 \cdot P_{cu,\text{FL}} = 162 \] \[ 0.81 \cdot P_{cu,\text{FL}} = 162 \] \[ P_{cu,\text{FL}} = \frac{162}{0.81} = 200 \text{ W} \]

Step 2: Compute the losses at 80% load condition ($x_2 = 0.8$). Now, let us find the individual losses when the transformer operates at a fractional loading of $x_2 = 0.8$:
• The iron loss remains unchanged because it is independent of the load: \[ P_i = 162 \text{ W} \]
• The copper loss at 80% load is recalculated using the full-load copper loss: \[ P_{cu}(0.8) = (0.8)^2 \cdot P_{cu,\text{FL}} = 0.64 \times 200 \text{ W} = 128 \text{ W} \]

Step 3: Calculate the total combined loss. Summing both the constant iron loss and the new variable copper loss together at 80% load: \[ P_{\text{total}}(0.8) = P_i + P_{cu}(0.8) = 162 \text{ W} + 128 \text{ W} = 290 \text{ W} \;\dots \text{ Wait, let's re-read carefully.} \] Let's carefully verify the phrasing: "The copper and iron losses of a 1-$\phi$ transformer are equal at 90% of full load and its value is 162 W." This phrasing can mean that the sum of the losses is 162 W, or that each individual loss is 162 W. Let's calculate both interpretations to verify which one matches the given multiple-choice options. Alternative Interpretation: Total loss at 90% load is 162 W. If the losses are equal and their total combined sum is 162 W: \[ P_i + P_{cu}(0.9) = 162 \text{ W} \quad \Rightarrow \quad 2 \cdot P_i = 162 \text{ W} \quad \Rightarrow \quad P_i = 81 \text{ W} \] Then the copper loss at 90% load is also $P_{cu}(0.9) = 81 \text{ W}$. Let us find the full-load copper loss under this assumption: \[ 0.81 \cdot P_{cu,\text{FL}} = 81 \text{ W} \quad \Rightarrow \quad P_{cu,\text{FL}} = \frac{81}{0.81} = 100 \text{ W} \] Now, let us calculate the copper loss at 80% load ($x = 0.8$) using this value: \[ P_{cu}(0.8) = (0.8)^2 \cdot 100 \text{ W} = 0.64 \times 100 = 64 \text{ W} \] Now, adding the constant iron loss ($P_i = 81 \text{ W}$) to find the total loss at 80% load: \[ P_{\text{total}}(0.8) = P_i + P_{cu}(0.8) = 81 \text{ W} + 64 \text{ W} = 145 \text{ W} \] This calculated value of 145 W perfectly matches Option (1). Therefore, the problem statement meant that the combined sum of the copper and iron losses at 90% load is 162 W.