An inductor of inductance L = 400 mH and resistors of $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ are connected to a battery of emf 12 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 :

12 e^{- 5 t} V

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\frac{12}{t} e^{3 t} V

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6 (1 - e^{- \frac{t}{0.2}}) V

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6 e^{- 5 t} V

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Solution

The correct option is A $12 e^{- 5 t} V$
In the branch containing L and $R_{2}$
$i = \frac{E}{R_{2}} ⎛ ⎜ ⎝ 1 - e^{- \frac{R_{2} t}{L}} ⎞ ⎟ ⎠$
$\frac{d i}{d t} = \frac{E}{R_{2}} e^{- \frac{R_{2} t}{L}} \cdot \frac{R_{2}}{L} = \frac{E}{L} e^{- \frac{R_{2} t}{L}}$
$∴ V_{L} = L \frac{d i}{d t} = E e^{- \frac{R_{2} t}{L}}$
$= 12 e^{- \frac{2 t}{400 \times 10^{- 3}}} = 12 e^{- 5 t} V$

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Similar questions

Q. An inductor of inductance

L = 400 m H

and resistors of resistance

R_{1} = 2 Ω, R_{2} = 2 Ω

are connected to a battery of e.m.f

E = 12 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

An inductor of inductance L = 400 mH and resistors of resistance $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ areconnected to a battery of emf 12 V as shown in the figure. Theinternal 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