Physics - EMI/AC Question with Solution | TestHub

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$.