An inductor $\left(L=0.03\text{H}\right)$ and a resistor $\left(R=0.15\text{kΩ}\right)$ 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 : $e^5\cong 150$

• Correct answer
  A

  0.67 mA
• B

  100 mA
• C

  67 mA
• D

  6.7 mA

Solution

Case I: ${\text{K}}_1$ is closed for long tiime



for long time, inductor acts as a conducting wire.

$\Rightarrow$  current in the circuit $=\frac{\text{V}}{\text{R}}$$=\frac{15}{150}$

${\text{i}}_0=\text{0.1 A}$

Case II: ${\text{K}}_1$ is open and ${\text{K}}_2$ is closed



Current in the circuit

$\text{i}={\text{i}}_0{\text{e}}^{\frac{-\text{t}}{\tau }}\text{; }\tau =\frac{\text{L}}{\text{R}}$

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

$\text{i}={\text{i}}_0{\text{e}}^{-\left(\frac{10^{-3}\times 150}{3\times 10^{-2}}\right)}$

$=\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'.