JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A transformer consisting of \(300\) turns in the primary and \(150\) turns in the secondary gives output power of \(2.2\, kW\). If the current in the secondary coils is \(10\, A\), then the input voltage and current in the primary coil are
- A \(440\, V\) and \(5\, A\)
- B \(440\, V\) and \(20\, A\)
- C \(220\, V\) and \(20\, A\)
- D \(220\, V\) and \(10\, A\)
Answer & Solution
Correct Answer
(A) \(440\, V\) and \(5\, A\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{N}_{\mathrm{P}}=300, \mathrm{N}_{\mathrm{s}}=150, \mathrm{P}_{0}=2200 \,\mathrm{W}\) \(\mathrm{l}_{\mathrm{s}}=10\, \mathrm{A}\) \(\mathrm{P}_{0}=\mathrm{V}_{0} \mathrm{l}_{0} \) \(\Rightarrow 2200=\mathrm{V}_{0} \times 10 \)…
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