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  • Need clarity, kindly explain! - Electromagnetic Induction and Alternating currents - JEE Main-4

 An inductor (L=0.03H) and a resistor   (R = 0.15 k\Omega) are connected in series to a battery of 15V EMF in a circuit shown below.  The key K1 has been kept closed for a long time. Then at t=0, K1 is opened and key K2 is closed simultaneously.At t=1ms, the current in the circuit will  \left ( e^{5}\cong 150 \right )

  • Option 1)

    100 mA

  • Option 2)

    67 mA

  • Option 3)

    6.7 mA

  • Option 4)

    0.67 mA

 

Answers (1)

 

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

 

 Initially at t = 0

I =     I_{o} = \frac{E}{R} = \frac{15}{150} A = 0.1 A

After closing the key K_{2}

I = I_{o} . e ^{- \frac{tR}{L}}

= (0.1) (e ^{- \frac{10 ^{-3} \times 150}{0.03}})

= 0.1 . e ^{-5} = \frac{0.1}{150} A

I = 0.67 mA

 


Option 1)

100 mA

This option is incorrect.

Option 2)

67 mA

This option is incorrect.

Option 3)

6.7 mA

This option is incorrect.

Option 4)

0.67 mA

This option is correct.

Posted by

Vakul

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