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 $\left(I_{\max }/I_{\min }\right)$ drawn from the battery is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (C)

When $t=0$ due to large impedance of two inductor current will flow only in $12\Omega$.
$\therefore I_{\min }=5/12$.
After sometime current become is steady then $R=12\Omega$ will go out of circuit only $r_1$ and $r_2$ will be effective route of current flow.
$r_{\text{ eff}}=2\Omega$
$\Rightarrow I_{\text{ max}}=\frac{5}{2}$
$\Rightarrow \frac{I_{\text{max}}}{I_{\min }}=6$