Question

The steady-state current through $L_1$ after the switch has been closed for a very long time is

• A. $\frac{V_0}{R}$
• B. $\frac{V_0{L}_1}{R\left(L_1+L_2\right)}$
• C. $\frac{V_0{L}_2}{R\left(L_1+L_2\right)}$
• D. None of these

Answer 1

Answer: (C)

After a very long time $\left(t\to \infty \right)$ ,
$I_0=\frac{V}{R}$ ....(i)
Since the inductors are in parallel,
$L_{\text{1}}\frac{\text{d}{\text{I}}_{\text{1}}}{\text{d}\text{t}}=L_{\text{2}}\frac{\text{d}{\text{I}}_{\text{2}}}{\text{d}\text{t}}$
After a long time,
$L_{\text{1}}{I}_{\text{1}}=L_{\text{2}}{I}_{\text{2}}$ .... (ii)
The total current $I_{\text{0}}$ through the battery is divided into the inductors, hence
$I_0=I_1+I_2$ ..... (iii)
From equations (ii) & (iii)
$I_1=\left(\frac{L_2}{L_1+L_2}\right)I_0\&I_2=\left(\frac{L_1}{L_1+L_2}\right)I_0$
The current through $L_{\text{1}}$ is
$I_1=\left(\frac{L_2}{L_1+L_2}\right)I_0$
$=\left(\frac{L_2}{L_1+L_2}\right)\frac{V_0}{R}$