Question

Two inductors $L_{1}$ (inductance $1\, mH$, internal resistance $3\, \Omega$ ) and $L_{2}$ (inductance $2 mH$, internal resistance $4\, \Omega$ ), and a resistor $R$ (resistance $12\, \Omega$ ) are all connected in parallel across a $5\, V$ 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. 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.
$r_{\text{ eff} }=2\, \Omega$
$\Rightarrow I_{\text{ max} }=\frac{5}{2}$
$\Rightarrow \frac{I_{ \text{max} }}{I_{\min }}=6$