# An inductor of inductance $\dpi{100} L=400\: mH$ and resistors of resistances $\dpi{100} R_{1}=2\, \Omega \; and\; R_{2}=2\, \Omega$ are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch $\dpi{100} S$ is closed at $\dpi{100} t=0$ . The potential drop across $\dpi{100} L$ as a function of time is Option 1) $6e^{-5t}V\; \;$ Option 2) $\; \frac{12}{t}e^{-3t}V\;$ Option 3) $\; \; 6(1-e^{-t/0.2})V\;$ Option 4) $\; 12e^{-5t}V$

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

time constant of the given circuit is$T=\frac{\mathrm{L} }{\mathrm{R} _{2}}=\frac{400mh}{2}$=0.2 sec

$\therefore$ potential drop is

$\xi =\varepsilon _{0}\cdot e^{-}\frac{\mathrm{t} }{\mathrm{T} } \\\xi =12\cdot e^{-5t}$

Option 1)

$6e^{-5t}V\; \;$

This is incorrect

Option 2)

$\; \frac{12}{t}e^{-3t}V\;$

This option is incorrect

Option 3)

$\; \; 6(1-e^{-t/0.2})V\;$

This option is incorrect

Option 4)

$\; 12e^{-5t}V$

This option is correct

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