Demonstrative Video


Efficiency of a transformer

\[\begin{aligned} \eta & =\dfrac{\mbox{output power}}{\mbox{input power}}=\dfrac{\mbox{output power}}{\mbox{output power+losses}}\\ & =\dfrac{\mbox{output power}}{\mbox{output power+iron losses+copper losses}}\\ & =\dfrac{V_{2}I_{2}\cos\Phi_{2}}{V_{2}I_{2}\cos\Phi_{2}+P_{i}+P_{c}} \end{aligned}\]
where
\[\begin{array}{cc} V_{2} & \mbox{Secondary terminal voltage}\\ I_{2} & \mbox{Full-load secondary current}\\ \cos\Phi_{2} & \mbox{Power factor of the load}\\ P_{i} & \mbox{Iron losses =hysterises +eddy current losses}\\ P_{c} & \mbox{Full load copper losses = }I^{2}R_{T} \end{array}\]
If \(x\) is the fraction of the full-load, then efficiency is given as
\[\eta_{x}=\dfrac{x\times\mbox{Output}}{x\times\mbox{Output}+P_{i}+x^{2}P_{c}}=\dfrac{xV_{2}I_{2}\cos\Phi_{2}}{xV_{2}I_{2}\cos\Phi_{2}+P_{i}+x^{2}I_{2}^{2}R_{T}}\]

Condition for the maximum efficiency

\[\eta = \dfrac{V_2I_2\cos\Phi_2}{V_2I_2\cos\Phi_2+P_i+I_2^2R_T} = \dfrac{V_2\cos\Phi_2}{V_2\cos\Phi_2+P_i/I_2+I_2R_T}\]
For maximum efficiency . Then is the fraction of full load KVA at which If . Hence, efficiency will be maximum when depends upon , is constant. Thus for a given Now,

All-Day Efficiency

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