MHT CET · Physics · Alternating Current
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
- A \(13.5 \mathrm{~A}, 225 \mathrm{~V}\)
- B \(13.5 \mathrm{~A}, \quad 100 \mathrm{~V}\)
- C \(4.5 \mathrm{~A}, \quad 100 \mathrm{~V}\)
- D \(6.0 \mathrm{~A}, \quad 225 \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(6.0 \mathrm{~A}, \quad 225 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\(6.0 A, 225 V\)
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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