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Practically how we see Copper losses are equal to iron losses in a transformer

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    \$\begingroup\$ In practice, how do we achieve equal copper and iron losses in a transformer? \$\endgroup\$ – DKNguyen Feb 3 at 5:54
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    \$\begingroup\$ Welcome to EE.SE. Please add some more words to your question, as it is hardly intellegible. \$\endgroup\$ – Ariser - reinstate Monica Feb 3 at 13:20
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We don't see iron losses equal to copper losses in a transformer.

Iron losses are due to the input voltage, so are the same (roughly) at no load and full load. In fact, they're slightly lower at full load as the primary voltage drop requires less flux swing.

Copper losses are due to the current flowing, so are small at no load and large at full load.

As the load varies, then so does the proportion of iron and copper losses.

When we design a transformer, we pick an operating level, and change the transformer dimensions to optimise losses at that operating point. Choosing to make iron and copper losses equal at that power level is a good starting point for optimising the transformer, but may not be the end of the optimisation.

For instance, when we make a transformer that spends most of its time at low load, like an audio power amplifier, we may choose a low flux to minimise iron hysteresis loss, maximising its inductance which minimises magnetising current, which minimises copper losses. The cost of this is that the many turns required means copper losses are high during its brief full load operation.

In contrast, when designing a microwave oven transformer which only operates at full load and has forced air cooling, flux is pushed up to the limit, and beyond, resulting in a magnetising current more or less equal in magnitude to the load current. However, these currents are in quadrature, so do not add up as badly as it first seems. This minimises the number of turns needed. These transformers get nearly as hot operated off load as on.

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Iron losses are only equal to copper losses at one operating point, which has the importance of being the point of maximum efficiency.

To understand this you have to know what iron and copper losses are, and how they vary with loading. Iron losses are caused by hysteresis and eddy currents in the magnetic core. For a given input voltage and frequency this loss is practically fixed (and usually quite small). Copper loss is caused by current flowing through the resistance of the windings. It is proportional to current squared, so it increases exponentially as output current increases.

With no load the transformer's efficiency is zero, since it uses some power to magnetize the core but outputs no power. As load is increased the efficiency goes up because it is now outputting power, but copper loss also increases so at some point the efficiency starts dropping again. The way the math works out, peak efficiency occurs when copper loss equals iron loss.

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