Question

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

• A. $4:1$
• B. $8:1$
• C. $6:1$
• D. $16:1$

Answer 1

The correct option is B $8:1$

At $t=0$ instant, all the inductors will act like an open branch. So no flow of current across them

$\therefore i_{min}=\frac{V}{R}$

Given:-
$R_1=3\Omega L_1=1\text{mH}$

$R_2=4\Omega L_2=2\text{mH}$

$R_3=12\Omega$

$\therefore i_{min}=\frac{V}{R_3}=\frac{5}{12}\text{A}$

At steady state, all the inductors will act as perfect conductor. So

$i_{max}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}$

$=5[\frac{1}{3}+\frac{1}{4}+\frac{1}{12}]=\frac{5(4+3+1)}{12}$

$\therefore i_{max}=\frac{40}{12}\text{A}$

$\therefore \frac{i_{max}}{i_{min}}=\frac{40}{12}\times \frac{12}{5}=8:1$

Hence, option $\left(\text{B}\right)$ is correct.