Physics - EMI/AC Question with Solution | TestHub

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$