Question

Two inductors $L_1$(inductance $1mH$, internal resistance $3\Omega$) and $L_2$(inductance $2mH$, internal resistance $4\Omega$), and a resistor R (resistance $12\Omega$) are all connected in parallel across a $5V$ battery. The circuit is switched on at time $t=0$. The ratio of the maximum to the minimum current ($I_{max}/I_{min}$) drawn from the battery is

Answer 1

As soon as the circuit is switched ON inductor behaves like open circuit due to high impedence of the inductors and hence only $12\Omega$ resistor will be in the closed circuit.
$i_{min}=\frac{V}{R}=\frac{5}{12}$ at $t=0^{+}$
At steady state i.e at $t=\infty$ inductor behaves like short circuit. So the equivalent resistance can be calculated for the parallel connection of the three resistors and that would be $\frac{12}{8}\Omega$.
$i_{max}=\frac{V}{R}=\frac{5\times 8}{12}$ at $t=\infty$
So the ratio
$\frac{i_{max}}{i_{min}}=8$