QJEEAdv16Phd767: JEE Advanced 2016 Physics - Class 12 JEEAdv16Phd767 Solution: Two inductors L 1 (inductance 1mH, internal resistance 3 ) and…

The circuit is shown in the figure

Initially, right after the circuit is switched on ( $t \rightarrow 0^{+}$ ), the impedance (effective resistance) of inductors $L_1$ and $L_2$ is extremely high. Therefore, the inductors act as open circuits, and all the current flows through the resistor $R = 12 \Omega$ . According to Kirchhoff's law, the current through the battery at this moment is $i_{\min} = \frac{V}{R} = \frac{5}{12} \, \text{A}$ .

As the circuit reaches steady state ( $t \rightarrow \infty$ ), the impedance of the inductors drops to zero, and they function as resistors with their specified internal resistances.

In the steady state, the current through the circuit is $i_{\max} = \frac{V}{R_e} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \, \text{A}$ .

Therefore, the ratio of the maximum current to the minimum current is:

$\frac{i_{\max}}{i_{\min}} = \frac{\frac{10}{3}}{\frac{5}{12}} = \frac{10}{3} \times \frac{12}{5} = 8$