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.

Answer 1

at $t$
$i _{1}=\frac{\varepsilon}{ R _{1}}\left(1- e ^{-\frac{ R _{1} t }{L_{1}}}\right)$
$i _{1}=\frac{5}{3}\left(1- e ^{-3 l }\right)$
$i _{2}=\frac{5}{4}\left(1- e ^{-2 x }\right)$
$i _{3}=\frac{5}{12}$
$i = i _{1}+ i _{2}+ i _{3}$
$i =\frac{5}{3}-\frac{5}{3} e ^{-3 t }+\frac{5}{4}-\frac{5}{4} e ^{-2 t}+\frac{5}{12}$
$i =\frac{10}{3}-\frac{5}{3} e ^{-3 t}-\frac{5}{4} e ^{-2 t }$
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 $t=\infty, i \rightarrow$ maximum
$i_{\max }=\frac{10}{3}$
Hence $\frac{I_{m a x}}{1_{\min }}=\frac{10}{3} \times \frac{12}{5}=8$