Correct Answer - Option 2 :

0.7A,155.6 W

Core loss α core volume

∴ New loss \({\left( {\sqrt 2 } \right)^3} \times 55 = 155.6W\)

Core loss \({P_i} = V{I_c}\)

\(\begin{array}{l}
{I_{{c_1}}} = \frac{{55}}{{1000}} = 0.055\ A\\
{I_{{c_2}}} = {\left( {\sqrt 2 } \right)^3}{I_{{c_1}}}
\end{array}\)

From the data for 2^{nd} transformer

\(\begin{array}{l}
{I_{{c_2}}} = 0.155\ A\\
\therefore {I_{{m_1}}} = 0.496\left[ {\therefore \sqrt {{{\left( {0.5} \right)}^2} - {{\left( {0.055} \right)}^2}} = 0.496} \right]
\end{array}\)

Reluctance \({R_2} = \frac{{{R_1}}}{{\sqrt 2 }}\)

\({\emptyset _{{m_1}}} = \frac{{4000}}{{4.44{N_1}f}}\) \({\emptyset _{{m_2}}} = \frac{{2000}}{{4.44\ {N_1}f}}\)

\(\begin{array}{l}
{\emptyset _{{m_2}}} = 2{\emptyset _{{m_1}}}\\
{\emptyset _{{m_1}}} = \frac{{{N_1}{I_m}_2}}{{{R_1}}}\\
{\emptyset _{{m_2}}} = \frac{{{N_1}{I_m}_2}}{{\frac{{{R_1}}}{{\surd 2}}}}\\
\therefore {I_{{m_2}}} = \surd 2{I_{{m_1}}}\\
= \sqrt 2 \times 0.496
\end{array}\)

= 0.702 A

\(\begin{array}{l}
\therefore {I_0} = \sqrt {I_{{m_1}}^2 + I_{{c_2}}^2} \\
= \sqrt {{{\left( {0.7025} \right)}^2} - {{\left( {0.155} \right)}^2}} = 0.718\ A
\end{array}\)