JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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 \(ln\,5 = 1.6\) ]
- A \(0.016\,s\)
- B \(0.324\,s\)
- C \(0.002\,s\)
- D \(0.103\,s\)
Answer & Solution
Correct Answer
(A) \(0.016\,s\)
Step-by-step Solution
Detailed explanation
\(\mathrm{L}=10 \times 10^{-3} \,\mathrm{H}, \mathrm{r}_{1}=0.1\, \Omega\) \(i=\varepsilon\left\{1-e^{-t / 2}\right\}\) \(i_{saturation}\) \(=\varepsilon\) \(80 \%\) \(i_{saturastion}\) \(=0.8\, \mathrm{\varepsilon}\) \(0.8 \varepsilon=\varepsilon\left\{1-e^{-t / 2}\right\}\)…
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