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  • Solve! - Electromagnetic Induction and Alternating currents - JEE Main

An inductor (L = 100 mH), a resistor (R = 100 \Omega ) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B . The current in the circuit 1 ms after the short circuit is

  • Option 1)

    1 A

  • Option 2)

    (1/e) A

  • Option 3)

    e A

  • Option 4)

    0.1 A

 

Answers (1)

As we learnt in 

Self Inductance -

An emf is induced in the coil or the circuit which oppose the change that causes it. Which is also known back  emf.

- wherein

Maximum current   I_{0}=\frac{E}{R}=\frac{100}{100}=1A

The current decays for 1 millisecon = 1 x 10-3 sec

During decay , I=I_{0}e^{-tR/L}

I=(1)e^{\frac{(-1\times 10^{-3})\times 100}{100\times 10^{-3}}}

or\; \; I=e^{-1}=\frac{1}{e}\ Amp

 


Option 1)

1 A

This option is incorrect

Option 2)

(1/e) A

This option is correct

Option 3)

e A

This is incorrect option

Option 4)

0.1 A

This option is incorrect

Posted by

Sabhrant Ambastha

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