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A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are
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The correct answer is:
$6.0 \mathrm{~A}, \quad 225 \mathrm{~V}$
$6.0 A, 225 V$
Given that, number of turns in primary winding, $N_{p}=300$
Number of turns in secondary winding, $N_{S}=450$
Primary voltage, $\mathrm{V}_{\mathrm{p}}=150 \mathrm{~V}$
Primary current, $I_{p}=9 \mathrm{~A}$
For step-up transformer,
$\begin{array}{l}
\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}} \\
\frac{V_{S}}{150}=\frac{450}{300} \\
\Rightarrow V_{S}=\frac{450}{300} \times 150 \\
\Rightarrow V_{S}=225 \mathrm{~V}
\end{array}$
$\begin{array}{l}
\text { Again, } \mathrm{V}_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}=\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{S}} \\
150 \times 9=225 \times \mathrm{I}_{\mathrm{S}} \\
\mathrm{I}_{\mathrm{s}}=\frac{1350}{225}=6.0 \mathrm{~A}
\end{array}$
Given that, number of turns in primary winding, $N_{p}=300$
Number of turns in secondary winding, $N_{S}=450$
Primary voltage, $\mathrm{V}_{\mathrm{p}}=150 \mathrm{~V}$
Primary current, $I_{p}=9 \mathrm{~A}$
For step-up transformer,
$\begin{array}{l}
\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}} \\
\frac{V_{S}}{150}=\frac{450}{300} \\
\Rightarrow V_{S}=\frac{450}{300} \times 150 \\
\Rightarrow V_{S}=225 \mathrm{~V}
\end{array}$
$\begin{array}{l}
\text { Again, } \mathrm{V}_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}=\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{S}} \\
150 \times 9=225 \times \mathrm{I}_{\mathrm{S}} \\
\mathrm{I}_{\mathrm{s}}=\frac{1350}{225}=6.0 \mathrm{~A}
\end{array}$
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