Find the current through the \$\mathrm{5 \space mH}\$ inductor when the circuit reaches a steady state
When the circuit reaches a steady state, a current of \$\mathrm{4 \space A}\$ will flow through the resistor (since voltage across the inductors are zero). The inductors themselves are ideal, and have a resistance of \$\mathrm{0 \space \Omega}\$. Thus, the current through each inductor should be \$\frac 42 = \mathrm{2 \space A}\$. However, my textbook seems to disagree and says that the current is \$\frac 83\$ A. I am aware that \$X_L = \omega L\$, but that is for an AC circuit and not a DC circuit.
Why is the current through two ideal inductors in parallel (at steady state) divided in the inverse ratio of their inductances?
Best Answer
The voltage between the ends of an ideal inductor: U=L*(di/dt) where term di/dt means the changing rate (=amperes/second) of the current through the inductor.
In practical inductors there's always some resistace and the equation would be U=L*(di/dt)+iR, but you declared R=0.
Your both inductors , say La=5mH and Lb=10mH have the same voltage, so
La*(d(ia)/dt)=Lb*(d(ib)/dt). That doesn't allow any other possibility that the current changing rates are inversely proportional to the inductances. Thus the cumulated currents are, too.