An ideal coil of 10 H is connected in series with a resistance of 5 \Omega and a battery of 5 V. 2 second after the connection is made, the current flowing in ampere in the circuit is

  • Option 1)

    (1-e^{-1})\;

  • Option 2)

    \; \; (1-e)\;

  • Option 3)

    \; \; e\; \;

  • Option 4)

    \; e^{-1}

 

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

During the growth of current in LR circuit is given by   I=I_{0}\left ( 1-e^{-\frac{R}{L}t} \right )

or\; \; I=\frac{E}{R}\left ( 1-e^{-\frac{R}{L}t} \right )=\frac{5}{5}\left ( 1-e^{-\frac{5}{10}\times 2} \right )

I=(I-e^{-1})

 


Option 1)

(1-e^{-1})\;

This option is correct

Option 2)

\; \; (1-e)\;

This option is incorrect

Option 3)

\; \; e\; \;

This option is incorrect

Option 4)

\; e^{-1}

This option is incorrect

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