Transformer losses are categorized based on their origin. The question asks which specific loss can be reduced by increasing the conductor's diameter.
Copper loss ($P_{cu}$) is calculated as:
$P_{cu} = I^2 R$
The resistance ($R$) of a conductor is given by:
$R = \frac{\rho L}{A}$
where:
The cross-sectional area ($A$) is directly related to the diameter ($d$) of the conductor. For a circular conductor, $A = \frac{\pi d^2}{4}$.
Therefore, to reduce the resistance ($R$), we need to increase the cross-sectional area ($A$). Increasing the diameter ($d$) of the conductor directly increases its cross-sectional area ($A$), which in turn decreases the resistance ($R$).
A lower resistance ($R$) leads to a reduction in copper loss ($P_{cu} = I^2 R$), assuming the current ($I$) remains constant.
By increasing the diameter of the conductor used for the windings, the cross-sectional area increases, resistance decreases, and consequently, the copper loss ($I^2R$) is reduced.
Copper loss is negligible at _____.