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  • Please help! - Electromagnetic Induction and Alternating currents - JEE Main

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 value is : \left [ taken \, \, \ln 5=1.6 \right ]

 

  • Option 1)

    0.324s

  • Option 2)

    0.103s

  • Option 3)

    0.002s

  • Option 4)

    0.016s

Answers (1)

best_answer

i=i_{0}\left ( 1-e^{-\frac{t}{\tau }} \right )

\frac{80}{100}i_{0}=i_{0}\left ( 1-e^{-\frac{t}{\tau }} \right )

\Rightarrow 0.8=1-e^{-\frac{t}{\tau }}

\Rightarrow e^{-\frac{t}{\tau }}=0.2

\Rightarrow -\frac{t}{\tau }=\ln \left ( 0.2 \right )

\Rightarrow t=\tau \ln 5=t=\frac{L}{R _{eq} }\ln 5

=\frac{10\times 10^{-3}}{\left ( 0.1+0.9 \right )}\times 1.6

\Rightarrow t= 0.016second


Option 1)

0.324s

Option 2)

0.103s

Option 3)

0.002s

Option 4)

0.016s
Posted by

Aadil

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