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…
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)
0.8 A
1.2 A
1.5 A
1.8 A
Solution
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,
$\frac{I_s}{I_p}=\frac{N_p}{N_s}$
$\mathrm{I}_{\mathrm{s}}=\frac{10^3}{2 \times 10^4} \times 30=1.5 \mathrm{~A}$