Question

Find the steady-state current through $L_{1}$ in the figure

• A. $\frac{V_{0}}{R}$
• B. $\frac{V_{0} L_{1}}{R \left(\right. L_{1} + L_{2} \left.\right)}$
• C. $\frac{V_{0} L_{2}}{R \left(\right. L_{1} + L_{2} \left.\right)}$
• D. none of these

Answer 1

Answer: (C)

In steady-state, both inductors will be short circuit and current drawn from the cell is,
$I_{0}=\frac{V_{0}}{R}$
This current will be divided in $L_{1}$ and $L_{2}$ like in parallel resistors
$I_{1}=I_{0}\frac{L_{2}}{L_{1} + L_{2}}$