Question:
In a single-phase transformer, total iron loss is 2500 W at 440 V, 50 Hz. At 220 V, 25 Hz, the iron loss is 850 W. Find hysteresis and eddy current losses at nominal voltage and frequency.
In a single-phase transformer, total iron loss is 2500 W at 440 V, 50 Hz. At 220 V, 25 Hz, the iron loss is 850 W. Find hysteresis and eddy current losses at nominal voltage and frequency.
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Hysteresis loss ∝ f, eddy current loss ∝ f² — very useful for transformer iron-loss calculations.
Updated On: Dec 29, 2025
- 1600 W and 900 W
- 900 W and 1600 W
- 250 W and 600 W
- 600 W and 250 W
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The Correct Option is B
Solution and Explanation
Iron loss = hysteresis loss + eddy current loss.
Hysteresis loss $\propto f$, eddy loss $\propto f^{2}$.
Let $P_h$ = hysteresis loss at 50 Hz, and $P_e$ = eddy current loss at 50 Hz.
At nominal (440 V, 50 Hz): \[ P_h + P_e = 2500. \]
At 220 V, 25 Hz: Voltage is halved → flux halves → eddy loss reduces by factor 4. Frequency halves → hysteresis loss halves.
Thus: \[ 0.5 P_h + 0.25 P_e = 850. \]
Solve the equations: 1) $P_h + P_e = 2500$ 2) $0.5 P_h + 0.25 P_e = 850$
Multiply (2) by 4: \[ 2P_h + P_e = 3400. \]
Subtract (1): \[ P_h = 900. \] Then \[ P_e = 2500 - 900 = 1600. \]
Final Answer: 900 W (hysteresis), 1600 W (eddy)
Hysteresis loss $\propto f$, eddy loss $\propto f^{2}$.
Let $P_h$ = hysteresis loss at 50 Hz, and $P_e$ = eddy current loss at 50 Hz.
At nominal (440 V, 50 Hz): \[ P_h + P_e = 2500. \]
At 220 V, 25 Hz: Voltage is halved → flux halves → eddy loss reduces by factor 4. Frequency halves → hysteresis loss halves.
Thus: \[ 0.5 P_h + 0.25 P_e = 850. \]
Solve the equations: 1) $P_h + P_e = 2500$ 2) $0.5 P_h + 0.25 P_e = 850$
Multiply (2) by 4: \[ 2P_h + P_e = 3400. \]
Subtract (1): \[ P_h = 900. \] Then \[ P_e = 2500 - 900 = 1600. \]
Final Answer: 900 W (hysteresis), 1600 W (eddy)
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