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 $\left(t \rightarrow \infty\right)$ ,
$I_{0}=\frac{V}{R}$ ....(i)
Since the inductors are in parallel,
$L_{\text{1}}\frac{\text{d} \textit{I}_{\text{1}}}{\text{d} \textit{t}}=L_{\text{2}}\frac{\text{d} \textit{I}_{\text{2}}}{\text{d} \textit{t}}$
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
$I_{1}=\left(\frac{L_{2}}{L_{1}+L_{2}}\right) I_{0}$
$=\left(\frac{L_{2}}{L_{1}+L_{2}}\right) \frac{V_{0}}{R}$