JEE Mains · Physics · STD 12 -6. Electromagnetic induction
In the given figure, an inductor and a resistor are connected in series with a battery of emf \(E\) volt. \(\frac{E^a}{2 b }\,J / s\) represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of \(\frac{ b }{ a }\) will be ............
- A \(24\)
- B \(23\)
- C \(25\)
- D \(22\)
Answer & Solution
Correct Answer
(C) \(25\)
Step-by-step Solution
Detailed explanation
\(E=\frac{1}{2} L I^2\) Rate of energy storing \(=\frac{d E}{d t}=L I \frac{d I}{d t}\) Now we Know for \(R-L\) circuit \(I=\frac{E}{R}\left(1-e^{-t \frac{R}{L}}\right)\) So \(\frac{d I}{d t}=\frac{E}{L} e^{-\frac{t R}{L}}\)…
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