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  4. An inductor (. L = 0.03 textH .) and a resistor ( R = 0.15 k Ω ) are connected in series to a battery of 15 textV EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t=0 , K1 is opened and key K2 is closed simultaneously. At t=1 textms , the current in the circuit will be: e5≅150 <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-rxzwr5gwfuv0.png />

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Q. An inductor $\left(\right. L = 0.03 \, \text{H} \left.\right)$ and a resistor $( R = 0.15 \, {k \Omega } )$ are connected in series to a battery of $15 \, \text{V}$ EMF in a circuit shown below. The key $K_{1}$ has been kept closed for a long time. Then at $t=0$ , $K_{1}$ is opened and key $K_{2}$ is closed simultaneously. At $t=1 \, \text{ms}$ , the current in the circuit will be : $e^{5}\cong150$

Question

NTA AbhyasNTA Abhyas 2020

Solution:

Case I: $\text{K}_{1}$ is closed for long tiime
Solution
for long time, inductor acts as a conducting wire.
$\Rightarrow $ current in the circuit $= \frac{\text{V}}{\text{R}}$ $= \frac{1 5}{1 5 0}$
$\text{i}_{0} = \text{0.1 A}$
Case II: $\text{K}_{1}$ is open and $\text{K}_{2}$ is closed
Solution
Current in the circuit
$i = i _{0} e ^{\frac{- t }{\tau}} ; \tau=\frac{ L }{ R }$
After $t =1 ms =10^{-3} s$
$i = i _{0} e ^{-\left(\frac{10^{-3} \times 150}{3 \times 10^{-2}}\right)}$
$=0.1 e ^{\frac{-15}{3}}=0.1 \frac{1}{ e ^{5}}=\frac{0.1}{150} A =0.67 mA$