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  • Answer please! - Electromagnetic Induction and Alternating currents - JEE Main

A coil of inductance 300 mH and resistance 2\Omega is connected to a source of voltage 2 V. The current reaches half of its steady state value in

  • Option 1)

    0.15 s

  • Option 2)

    0.3 s

  • Option 3)

    0.05 s

  • Option 4)

    0.1 s

 

Answers (1)

best_answer

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

During growth of charge in an inductance,

I=I_{0}(1-e^{-Rt/L})

or \; \; \frac{I_{0}}{2}=I_{0}(1-e^{-Rt/L})

or\; \; e^{-Rt/L}=\frac{1}{2}=2^{-1}

or\; \; \frac{Rt}{L}=1n2\Rightarrow t=\frac{L}{R}1n2

t=\frac{300\times 10^{-3}}{2}\times (0.693)

or\; \; \; t=0.1 sec

 


Option 1)

0.15 s

Incorrect

Option 2)

0.3 s

Incorrect

Option 3)

0.05 s

Incorrect

Option 4)

0.1 s

Correct

Posted by

Aadil

View full answer

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