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

6 e^{- 5 t} V

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\frac{12}{7} e^{- 3 t} V

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6 ⎡ ⎢ ⎢ ⎣ 1 - e^{(- \frac{t}{0.2})} ⎤ ⎥ ⎥ ⎦ V

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12 e^{- 5 t} V

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Solution

The correct option is D $12 e^{- 5 t} V$
Let
$I_{1}$ be current through the resistor $R_{1}$
$I_{2}$ be current through the inductor

Using Kirchhoffs voltage law,
$I_{1} = \frac{E}{R_{1}} = \frac{12}{2} = 6 A$ and $E = L \frac{d I_{2}}{d t} + R_{2} I_{2}$

$\Rightarrow I_{2} = I_{0} ⎡ ⎢ ⎢ ⎣ 1 - e^{- \frac{t}{T_{c}}} ⎤ ⎥ ⎥ ⎦$
As steady state current $I_{0} = \frac{E}{R_{2}} = \frac{12}{2} = 6 A$
Time constant of the circuit $T_{c} = \frac{L}{R} = \frac{400 \times 10^{- 3}}{2} = 0.2 s$
$\Rightarrow I_{2} = 6 ⎡ ⎢ ⎣ 1 - e^{- \frac{t}{0.2}} ⎤ ⎥ ⎦$

Potential drop across $L$ is $= E - R_{2} I_{2} = 12 - 2 \times 6 [1 - e^{- 5 t}] = 12 e^{- 5 t} V$

Suggest Corrections

Similar questions

Q. An inductor of inductance

L = 400 m H

and resistors of resistances

R_{1} = 2 Ω

and

R_{2} = 2 Ω

are connected to a battery of emf

12 V

as shown in 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