Question 142 of 190

An inductor $(L=0.03 \text{H})$ and a resistor $(R=0.15 \text{kΩ})$ are connected in series to a battery of $15 \text{V}$ E.M.F. 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 : $\left(\text{Take},{e}^{5}\approx 150\right)$

A$0.67mA$B$100mA$C$67mA$D$6.7mA$

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Answer: (A) $0.67mA$

Case I: ${\text{K}}_{1}$ is closed for long time,



for a long time, the inductor acts as a conducting wire, $\Rightarrow$current in the circuit $=\dfrac{\text{V}}{\text{R}}$$=\dfrac{15}{150}$$\text{⇒}{\text{i}}_{0}=\text{0.1 A}$
Case II: ${\text{K}}_{1}$ is open and ${\text{K}}_{2}$ is closed



Current in the circuit, $\text{i}={\text{i}}_{0}{\text{e}}^{\dfrac{-\text{t}}{\tau }}\text{;}\tau =\dfrac{\text{L}}{\text{R}}$

After $\text{t}=1\text{ms}=1{0}^{-3}\text{s}$

$\text{⇒}\text{i}={\text{i}}_{0}{\text{e}}^{-\left(\dfrac{1{0}^{-3}\times 150}{3\times 1{0}^{-2}}\right)}$$=\text{0.1}{\text{e}}^{\dfrac{-15}{3}}$$=\text{0.1}\dfrac{1}{{\text{e}}^{5}}=\dfrac{\text{0.1}}{150}\text{A}$$=\text{0.67}mA$.