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.

Answer 1

Answer: 8

at $t$
$i_1=\frac{\epsilon }{R_1}\left(1-e^{-\frac{R_{1}t}{L_1}}\right)$
$i_1=\frac{5}{3}\left(1-e^{-3l}\right)$
$i_2=\frac{5}{4}\left(1-e^{-2x}\right)$
$i_3=\frac{5}{12}$
$i=i_1+i_2+i_3$
$i=\frac{5}{3}-\frac{5}{3}{e}^{-3t}+\frac{5}{4}-\frac{5}{4}{e}^{-2t}+\frac{5}{12}$
$i=\frac{10}{3}-\frac{5}{3}{e}^{-3t}-\frac{5}{4}{e}^{-2t}$
at $t=0$
Current (i) would be minimum
at $t=0$
$i_{\min }=\frac{10}{3}-\frac{5}{3}-\frac{5}{4}=\frac{5}{12}$
at $t=\infty ,i\to$ maximum
$i_{\max }=\frac{10}{3}$
Hence $\frac{I_{max}}{1_{\min }}=\frac{10}{3}\times \frac{12}{5}=8$