An inductor of inductance L=400 mH and resistors of resistance R_ 1 =2 and R_ 2 =2 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

Correct answer: C. 12 ^ -5 t V Growth in current in L R_ 2 branch when switch is closed is given by array l i= E R_ 2 [1-e^ -R_ 2 t / L ] (d i)/(d t)= E R_ 2 · R_ 2 L e^ -R_ 2 t / L =(E)/(L)e^ - R_ 2 t L array Hence, potential drop across L=((E)/(L) e^ -R_ 2 t / L ) L=E e^ -R_ 2 t / L =12 e^ - 2 t 400 × 10^ -3 =12 ^ -5 t ~V

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Electromagnetic Induction

BITSAT 2016 · Physics · Electromagnetic Induction

An inductor of inductance L=400 mH and resistors of resistance R_ 1...

An inductor of inductance

L=400 \mathrm{mH}

and resistors of resistance

R_{1}=2 \Omega

and

R_{2}=2 \Omega

are connected to a battery of emf

12 \mathrm{ 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 and R_ 2 =2 are connected to a battery of emf 12

Options

A $\frac{12}{t} e^{-3 t} \mathrm{ V}$
B $6\left(1-e^{-t / 0.2}\right) \mathrm{V}$
C $12 \mathrme^{-5 t} \mathrm{ V}$
D $6 \mathrme^{-5 t} \mathrm{ V}$

Correct answer

C. $12 \mathrme^{-5 t} \mathrm{ V}$

Step-by-step solution

Growth in current in

L R_{2}

branch when switch is closed is given by

\begin{array}{l}i=\frac{E}{R_{2}}\left[1-e^{-R_{2} t / L}\right] \\ \Rightarrow \frac{d i}{d t}=\frac{E}{R_{2}} \cdot \frac{R_{2}}{L}e^{-R_{2} t / L}=\frac{E}{L}e^{-\frac{R_{2} t}{L}} \end{array}

Hence, potential drop across $\mathbf{L}=\left(\frac{E}{L} e^{-R_{2} t / L}\right) L=E e^{-R_{2} t / L}=12 e^{-\frac{2 t}{400 \times 10^{-3}}}=12 \mathrme^{-5 t} \mathrm{~V}

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