Question

Two coils $A$ and $B$ have inductance $1.0H$ and $2.0H$ respectively. The resistance of each coil is $10\Omega$. Each coil is connected to an ideal battery of emf $2.0V$ at $t=0$. Let $i_A$ and $i_B$ be the currents in the two circuit at time $t$. Find the ratio $i_A/i_B$ at (a) $t=100ms$, (b) $t=200ms$ and (c) $t=1s$

Answer 1

$L_A=1.0H;L_B=2.0H;R=10\Omega t=0.1s,{\tau }_A=0.1,{\tau }_B=L/R=0.2$
$i_A=i_0(1-e^{-t/\tau })=\frac{2}{10}(1-e^{\frac{-0.1\times 10}{1}})=0.2(1-e^{-1})=0.126424111$
$i_B=i_0(1-e^{-t/\tau })=\frac{2}{10}(1-e^{\frac{-0.1\times 10}{2}})=0.2(1-e^{-1/2})=0.078693$
$\frac{i_A}{i_B}=\frac{0.12642411}{0.78693}=1.6$
(b) $t=200ms=0.2s$
$i_A=i_0(1-e^{-t/\tau })=0.2(1-e^{-0.2\times 10/1})=0.2\times 0.864663716=0.172932943$
$i_B=i_0(1-e^{-t/\tau })=0.2(1-e^{-0.2\times 10/2})=0.2\times 0.632120=0.126424111$
$\frac{i_A}{i_B}=\frac{0.172932943}{0.126424111}=1.36=1.4$
(c) $t=1s$
$i_A=0.2(1-e^{-1\times 10/1})$
$i_B=0.2(1-e^{-1\times 10/2})=0.2\times 0.99326=0.19865241$
$\frac{i_A}{i_B}=\frac{0.19999092}{0.19865241}=1.0$