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# In the given circuit, the current through the 5mH inductor in steady state is $\left( {\text{A}} \right){\text{ }}\dfrac{2}{3}A$ $\left( {\text{B}} \right){\text{ }}\dfrac{8}{3}A$ $\left( {\text{C}} \right){\text{ }}\dfrac{1}{3}A$ $\left( {\text{D}} \right){\text{ }}\dfrac{5}{3}A$

Last updated date: 01st Mar 2024
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Hint: The inductors are connected in parallel; think which quantities will be the same for this combination and which quantities would change as the current flows in the circuit.

Formulae used:
$V = L\dfrac{{di}}{{dt}}$ , $I = \dfrac{V}{R}$ here, $I$ is current, $R$ is resistance and $V$ is potential difference.

Complete step by step solution:
The steady current that flows in the circuit is given by-
$I = \dfrac{V}{R}$
From the figure, $V = 20V$ and $R = 5\Omega$
$\therefore I = \dfrac{{20}}{5} = 4A$
Now, when the current passes through the inductors, current $I$ splits into ${i_1}$ (across $5{\text{ }}mH$ inductor) and i2 (across $10mH$ inductor)
$\therefore I = {i_1} + {i_2}$ ---------(1)
We need to find current across 5mH inductor i.e. $\;{i_2}$
Now since the inductors are connected in parallel then, the voltage across them will be same
$\therefore {V_1} = {V_2}$ (${V_1}$ for $5mH$ and ${V_2}$ for$10mH$)
$\Rightarrow {L_1}\dfrac{{d{i_1}}}{{dt}} = {L_2}\dfrac{{d{i_2}}}{{dt}}$
After integration,
${L_1}{i_1} = {L_2}{i_2}$ --------(2)
From equation 1,
${i_2} = I - {i_1}$
Substituting in equation 2, we get
${L_1}{i_1} = {L_2}(I - {i_1})$
$\Rightarrow ({L_1} + {L_2}){i_1} = {L_2}I$
$\therefore {i_1} = I\left( {\dfrac{{{L_2}}}{{{L_1} + {L_2}}}} \right)$
Substituting the values of ${L_1}$ and ${L_2}$ we get,
${i_1} = \left( {\dfrac{{10}}{{10 + 5}}} \right) = 4\left( {\dfrac{{10}}{{15}}} \right) = \dfrac{8}{3}A$