JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2018, 2019
- Medium
- Numerical
A power transmission line feeds input power at \(2300\,V\) to a step down transfonner with its primary windings having \(4000\) turns, giving the output power at \(230\,V.\) If the current in the primary of the transformer is \(5\,A,\) and its efficiency is \(90\%,\) the output current would be......\(A\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(45\)
Step-by-step Solution
Detailed explanation
Given \(: V_{P}=2300\, \mathrm{V}, V_{s}=230\, \mathrm{V}, I_{P}=5\, \mathrm{A}\)
\(\eta=90 \%=0.9\)
Efficiency \(\eta=0.9=\frac{P_{s}}{P_{p}} \Rightarrow P_{s}=0.9 P_{p}\)
\(\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}=0.9 \times \mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}} \quad(\because \mathrm{P}=\mathrm{VI})\)
\({I_s} = \frac{{0.9 \times 2300 \times 5}}{{230}} = 45{\mkern 1mu} {\text{A}}\)
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