Question

Two indicators $L_1$ (inductance $1\text{mH}$, internal resistance $6\Omega$) and $L_2$ (inductance $2\text{mH},$ internal resistance $6\Omega$), and a resistor $R$ (resistance $6\Omega$) are all connected parallel across a $12\text{V}$ battery. The circuit is switched on at time $t=0$. The ratio of the maximum to minimum current $\left(I_{max}/I_{min}\right)$ drawn from the battery is (only integer)

Answer 1

At $t=0$, inductance acts as a resistor of infinite resistance.

Hence, $R_{eff}=R_1=R=6\Omega$.

At steady state, inductors offer zero resistance, hence,

$\frac{1}{R_{eff}}=\frac{1}{R_2}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$

$\Rightarrow R_2=2\Omega$

Hence, the current will be maximum at steady state and minimum at $t=0$.

So, $I_{max}=\frac{E}{R_2}=\frac{12}{2}=6\text{A}$ and

$I_{min}=\frac{E}{R_1}=\frac{12}{6}=2\text{A}$

$\Rightarrow \frac{I_{max}}{I_{min}}=\frac{6}{2}=3$

Correct answer$:3$

Why this question ? Key trick: At the time when the current starts flowing, we will have a minimum current at $t=0$. In this time the whole current will pass through resistor only and no current would be passing through inductors. After a long time, current will also pass through inductors, and we will have our maximum current.