The coils of a step-down transformer have 500 and 5000 turns. In the primary coil an \(\mathrm{AC}\) current…
The coils of a step-down transformer have 500 and 5000 turns. In the primary coil an \(\mathrm{AC}\) current of \(4 \mathrm{~A}\) at \(2200 \mathrm{~V}\) is sent. The value of the current and potential difference in the secondary coil is
\(20 \mathrm{~A}, 220 \mathrm{~V}\)
\(0.4 \mathrm{~A}, 22000 \mathrm{~V}\)
\(40 \mathrm{~A}, 220 \mathrm{~V}\)
\(40 \mathrm{~A}, 22000 \mathrm{~V}\)
Solution
For a step-down transformer,
\(\begin{aligned}
N_P & =5000, N_S=500 \\
I_P & =4 \mathrm{~A}, V_P=2200 \mathrm{~V}
\end{aligned}\)
For a transformer, we know that
\(\begin{aligned}
& \frac{N_S}{N_P}=\frac{V_S}{V_P}=\frac{I_P}{I_S} \\
& \text {Using, } \quad \frac{N_S}{N_P}=\frac{V_S}{V_P} \\
& \Rightarrow \quad V_S=\frac{N_S V_P}{N_P}=\frac{500 \times 2200}{5000}=220 \mathrm{~V} \\
& \text {Again, using } \frac{V_S}{V_P}=\frac{I_P}{I_S} \\
& I_S=\frac{I_P \times V_P}{V_S}=\frac{4 \times 2200}{220}=40 \mathrm{~A} \\
\end{aligned}\)