An inductor (L=20 H) , a resistor (R=100 ) and a battery (E=10 ~V)...

An inductor

(L=20 \mathrm{H})

, a resistor

(R=100 \Omega)

and a battery

(E=10 \mathrm{~V})

are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1 ms after short circuiting.

Options

A $4.5 \times 10^5 \mathrm{~A}$
B $3.2 \times 10^{-5} \mathrm{~A}$
C $9.8 \times 10^{-5} \mathrm{~A}$
D $6.7 \times 10^{-4} \mathrm{~A}$

Correct answer

D. $6.7 \times 10^{-4} \mathrm{~A}$

Step-by-step solution

The initial current

i=i_0=\frac{E}{R}=\frac{10}{100} =0.10 \mathrm{~A}

the time constant,

\tau=\frac{L}{R}=\frac{20 \mathrm{mH}}{100 \Omega} =0.20 \mathrm{~ms}

The current at

t=1 \mathrm{~ms}

\begin{aligned} i & =i_0 e^{-t / \tau} \\ & =(0.10) e^{\left(\frac{-1 \mathrm{~ms}}{0.20 \mathrm{~ms}}\right)} \\ & =(0.10) e^5=6.7 \times 10^{-4} \mathrm{~A}\end{aligned}

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