An ideal coil of 10 H is connected in series with a resistance of 5 $\dpi{100} \Omega$ 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) $\; e^{-1}$

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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