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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2019
  • Hard
  • MCQ

A coil of self inductance \(10\,mH\) and resistance \(0.1\,\Omega \) is connected through a switch to a battery of internal resistance \(0.9\,\Omega \). After the switch is closed the time taken for the current to attain \(80\%\) of the saturation values is: [take \(ln\,5 = 1.6\) ]

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Answer & Solution

Correct Answer

(A) \(0.016\,s\)

Step-by-step Solution

Detailed explanation

\(\mathrm{L}=10 \times 10^{-3} \,\mathrm{H}, \mathrm{r}_{1}=0.1\, \Omega\)

\(i=\varepsilon\left\{1-e^{-t / 2}\right\}\)

\(i_{saturation}\) \(=\varepsilon\)

\(80 \%\) \(i_{saturastion}\) \(=0.8\, \mathrm{\varepsilon}\)

\(0.8 \varepsilon=\varepsilon\left\{1-e^{-t / 2}\right\}\)

\(0.8=1-\mathrm{e}^{-\mathrm{t/2} } \quad ; \quad \mathrm{e}^{-\mathrm{t/2} }=0.2\)

\(e^{t/L}=5\)

\(\mathrm{t}=\mathrm{L} \ln 5=10 \times 10^{-3} \times 1.6=16 \times 10^{-3}\)

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