Question

In the given circuit, find the ratio of $i_1$ to $i_2$ where $i_1$ is the initial (at t = 0) current and $i_2$ is steady state $(att=\infty )$ current through the battery.

Answer 1

Inductor is

1) open circuit at $t=0$ and

2) short circuited at $t=\infty$

Hence

$R_{eq}$ at $t=0$

$R_{eq}=6+4=10\Omega$

$\Rightarrow I_1=\frac{10V}{10R}=1A$..........(i)

Req at $t=\infty$

$R_{eq}=6+\left(\frac{4\times 4}{4+4}\right)=8\Omega$.

$\Rightarrow I_2=\frac{10V}{8R}=\frac{10}{8}A$..........(ii)

Now eq $\frac{i}{ii}$

$\frac{I_1}{I_2}=\frac{8}{10}=0.8$