A transformer is used to set up an alternating e.m.f. of 220 V to 4.4 kV to transmit 6.6 kW of power. The primary coil has 1000 turns. The current rating of the secondary coil is (Transformer is ideal)

Number of turns of in primary and secondary coil be \(\mathrm{N}_{\mathrm{p}}\) and \(\mathrm{N}_{\mathrm{s}}\),
For ideal transformer,
\(\begin{aligned}
& \frac{N_s}{N_p}=\frac{V_s}{V_p} \\
& N_s=\frac{4.4 \times 10^3}{220} \times 1000=2 \times 1.0^4
\end{aligned}\)
Power supply at primary coil is given as,
\(\begin{aligned}
& \mathrm{P}=\mathrm{I}_{\mathrm{p}} \mathrm{~V}_{\mathrm{p}}=6.6 \times 10^3 \mathrm{~V} \\
& \mathrm{I}_{\mathrm{p}}=\frac{6.6 \times 10^3 \mathrm{~V}}{220}=30 \mathrm{~A}
\end{aligned}\)
For ideal transformer,
\(\frac{I_s}{I_p}=\frac{N_p}{N_s}\)
\(\mathrm{I}_{\mathrm{s}}=\frac{10^3}{2 \times 10^4} \times 30=1.5 \mathrm{~A}\)