The core loss of a single phase, 230/115 V, 50Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage variable frequency source while keeping the secondary open circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is gain measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are respectively

For V = 230 V and F = 50 Hz, we have core LOSS = 1050 W or Iron loss + Hysteresis loss = 1050 WSince, iron loss and hysteresis loss are given by Iron loss \(= {P_i}\; = {k_i}\;{V^2}\) and Hysteresis loss \(= \;{P_H} = {K_H}\;{V^{1.6}}\;{f^{ - 0.6}}\)Substituting it in equation (i) we get \({K_i}{\left( {230} \right)^2} + {K_H}\frac{{{{\left( {230} \right)}^{1.6}}}}{{{{\left( {50} \right)}^{0.6}}}} = 1050\) or\({K_i}\left( {52900} \right) + {K_H}\left( {574.62} \right) = 1050\)Again for \(V\; = \;138\;V\) and \(f\; = \;30\;Hz\), we haveCore loss = 500 W or Iron loss + Hysteresis loss = 500 Wor \({K_P}{\left( {138} \right)^2} + {K_H}\frac{{{{\left( {138} \right)}^{1.6}}}}{{{{\left( {30} \right)}^{0.6}}}} = 500\) or KP (19044) + KH (344.77) = 500Solving equations (i) and (ii), we get\({K_P}\; = \;0.01027\;\;\;\;{K_H}\; = \;0.855\)And, For 230 V, 50Hz we haveHysteresis loss = \({K_H}\;{V^{1.6}}\;{f^{ - 0.6}}\) \(= 0.855 \times {\left( {230} \right)^{1.6}} \times {\left( {50} \right)^{ - 0.6}}\; = \;508.53\;W\)HENCE, the eddy loss is Eddy loss = ­\({K_{I{V^2}}} = {\text{}}0.01024 \times 529000 = 541.69\;W\)