An inductor of inductance L = 400 mH and resistors of resistance R₁ =… | JEE Main 2009 (AIEEE)

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Inductors & DC Transients (Mixed)

An inductor of inductance $L = 400\text{ mH}$ and resistors of resistance $R_1 = 2\ \Omega$ and $R_2 = 2\ \Omega$ are connected to a battery of $\text{emf } 12\text{ V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0.$ The potential drop across $L$ as a function of time is:

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Answer: (c)

$R_1$ and the $L$ - $R_2$ branch are each directly across the 12 V battery, so the $L$ - $R_2$ branch grows on its own.
Its time constant $\tau = \frac{L}{R_2} = \frac{0.4}{2} = 0.2\text{ s}$
Voltage across $L$ : $V_L = Ee^{-t/\tau} = 12e^{-t/0.2} = 12e^{-5t}\text{ V}$
So the answer is (c) $12e^{-5t}$ V.

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