question_answer

Two inductors \[{{L}_{1}}\] (inductance \[1\text{ }mH,\] internal resistance \[3\Omega \]) and \[{{L}_{2}}\] (inductance \[2\text{ }mH,\] 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 a time\[t=0\]. The ratio of the maximum to the minimum current \[({{I}_{\max }}/{{I}_{\min }})\]drawn from the battery is

A) 8         

B) 10     

C) 12       

D) 14

Correct Answer: A

Solution :

At \[t=0,\]current will flow only in \[12\Omega \] resistance

\[\therefore \,\,{{I}_{\min }}=\frac{5}{12}\]

At \[t\to \infty \] both \[{{L}_{1}}\]and \[_{2}\] behave as conducting wires

\[\therefore \,\,{{R}_{eff}}=\frac{3}{2},\,\,{{I}_{\max }}=\frac{10}{3}\]

\[\frac{{{I}_{\max }}}{{{I}_{\min }}}=8\]