An inductor $\left(L=0.03\text{H}\right)$ 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 : $\left(\text{Take}, e^5\approx 150\right)$

• Correct answer
  A

  $0.67 \mathrm{mA}$
• B

  $100 \mathrm{mA}$
• C

  $67 \mathrm{mA}$
• D

  $6.7 \mathrm{mA}$

Solution

Case I: ${\mathit{\text{K}}}_1$ is closed for long time, 



for a long time, the inductor acts as a conducting wire, $\Rightarrow$current in the circuit $=\frac{\mathit{\text{V}}}{\mathit{\text{R}}}$$=\frac{15}{150}$$\text{⇒}{\mathit{\text{i}}}_0=\text{0.1 A}$
Case II: ${\mathit{\text{K}}}_1$ is open and ${\mathit{\text{K}}}_2$ is closed



Current in the circuit, $\mathit{\text{i}}={\mathit{\text{i}}}_0{\text{e}}^{\frac{-\mathit{\text{t}}}{\tau }}\text{; }\tau =\frac{\mathit{\text{L}}}{\mathit{\text{R}}}$

After $\mathit{\text{t}}=1 \text{ms}=10^{-3} \text{s}$

$\text{⇒}\mathit{\text{i}}={\mathit{\text{i}}}_0{\text{e}}^{-\left(\frac{10^{-3}\times 150}{3\times 10^{-2}}\right)}$$=\text{0.1}{\text{e}}^{\frac{-15}{3}}$$=\text{0.1}\frac{1}{{\text{e}}^5}=\frac{\text{0.1}}{150} \text{A}$$=\text{0.67 }mA$.

Hence, the correct answer is option: 'A'.