Physics - EMI/AC Question with Solution | TestHub

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 $=\frac{\text{V}}{\text{R}}$ $=\frac{15}{150}$ \Rightarrow \textit{i}_{0} = \text{0}.\text{1 \mathrm{A}}Case II:

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

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

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

$\Rightarrow \text{i}={\text{i}}_0{\text{e}}^{-\left(\frac{10^{-3}\times 150}{3\times 10^{-2}}\right)}$ $=\text{0}.\text{1}{\text{e}}^{\frac{-15}{3}}$ $=\text{0}.\text{1}\frac{1}{{\text{e}}^5}=\frac{\text{0}.\text{1}}{150}\mathrm{A}$ $=\text{0}.\text{67 }\text{mA}$.