JEE Mains · Physics · STD 12 -7. Alternating current
A \(20\, Henry\) inductor coil is connected to a \(10\, ohm\) resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is
- A \(\frac{2}{{\ln \,2}}\)
- B \(ln\,2\)
- C \(\frac{1}{2}\,\ln\,2\)
- D \(2\,ln\,2\)
Answer & Solution
Correct Answer
(D) \(2\,ln\,2\)
Step-by-step Solution
Detailed explanation
\(L I D I=I^{2} R\) \(L \times \frac{E}{10}\left(-e^{-t / 2}\right) \times \frac{-1}{2}=\frac{E}{10}\left(1-e^{-t / 2}\right) \times 10\) \(e^{-1 / 2}=1-e^{-1 / 2} \quad ; \quad t=2 \ell n 2\)
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