The Question is:-
Find the current distribution in the inductors at steady state (i.e.
assuming $t\rightarrow \infty$ after the switch is closed and circuit
established and consider the inductors without any resistance and ideal).
My Efforts
Since both of the are in parallel, the equations for current at any time $t$ after establishing the circuit is given by ($i_1$=corrent in $5mH$ inductor and $i_2$ in the other, and the total current $i$):-
$$i(t)=i_1(t)+i_2(t)=4(1-e^{-\frac{5t}{L_{net}}})$$ where $L_{net}=\frac{10}{3}mH$. But I do not know how this current divides into two in the inductances at any time $t$ or in the steady state. Had the inductors not been ideal, i.e. if they had different resistances, the current could divide in the ration of their resistances in the final steady state, but since they are ideal, this method won't work. How can I solve this? the answers given are $\frac83A$ in the $50mH$ inductor and $\frac43A$ in the other.
Best Answer
Forget the distraction of the steady state specification for a moment. Since the inductors are parallel connected, it follows that:
$$L_1 \dfrac{di_1}{dt} = L_2 \dfrac{di_2}{dt}$$
Integrating both side with respect to time and assuming zero initial conditions yields:
$$L_1 i_1(t) = L_2 i_2(t)$$
Thus:
$$\dfrac{i_1(t)}{i_2(t)} = \dfrac{L_2}{L_1} = 2$$
And so,
$$i_1 = 2 \cdot i_2$$
for all $ t \ge 0$.
If there is series resistance (non-ideal inductors) then the analysis proceeds as follows:
$$L_1 \dfrac{di_1}{dt} + R_1 \cdot i_1 = L_2 \dfrac{di_2}{dt} + R_2 \cdot i_2$$
In steady state we have:
$$ \dfrac{di_1}{dt} = \dfrac{di_2}{dt} = 0$$
Thus:
$$R_1 \cdot i_1(\infty) = R_2 \cdot i_2(\infty) \rightarrow i_1(\infty) = \dfrac{R_2}{R_1} i_2(\infty)$$