# Current through an ideal inductor at steady state

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?

• Maybe they argue that when you turn on the circuit the current will first split according to the inductance and then the inductors will keep that current going. Jun 5, 2020 at 7:36
• Sometimes text books are wrong. Jan 29 at 19:57

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.

Well, let's solve this mathematically. We have the following circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1}{\text{R}_3} \end{cases}\tag2$$

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

$$\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1}{\text{R}_3}\tag3$$

Now, we can solve for $$\\text{V}_1\$$:

$$\text{V}_1=\frac{\text{V}_\text{i}}{1+\frac{\text{R}_1}{\text{R}_2}+\frac{\text{R}_1}{\text{R}_3}}\tag4$$

So, for $$\\text{I}_3\$$ we get:

$$\text{I}_3=\frac{\text{V}_1}{\text{R}_3}=\frac{\text{V}_\text{i}}{\text{R}_1+\text{R}_3+\frac{\text{R}_3\text{R}_1}{\text{R}_2}}=\frac{\text{V}_\text{i}\text{R}_2}{\text{R}_1\text{R}_2+\text{R}_2\text{R}_3+\text{R}_1\text{R}_3}\tag5$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

• $$\text{R}_2=\text{sL}_1\tag6$$
• $$\text{R}_3=\text{sL}_2\tag7$$

So, we get:

$$\text{i}_3\left(\text{s}\right)=\frac{\text{v}_\text{i}\left(\text{s}\right)\text{sL}_1}{\text{sL}_1\text{R}_1+\text{sL}_1\text{sL}_2+\text{sL}_2\text{R}_1}\tag8$$

Using the fact that $$\\text{V}_\text{i}\$$ is a stable DC-voltage, so we know that:

$$\text{v}_\text{i}\left(\text{s}\right)=\frac{\hat{\text{u}}_\text{i}}{\text{s}}\tag9$$

Where $$\\hat{\text{u}}_\text{i}\$$ is the value of the voltage source.

So, we can rewrite $$\(8)\$$:

$$\text{i}_3\left(\text{s}\right)=\frac{1}{\text{s}}\cdot\frac{\hat{\text{u}}_\text{i}\text{L}_1}{\text{s}\text{L}_1\text{L}_2+\text{L}_1\text{R}_1+\text{L}_2\text{R}_1}\tag{10}$$

Now, we can use the final value theorem of the Laplace transform to find:

$$\lim_{t\to\infty}\text{I}_3\left(t\right)=\lim_{\text{s}\to0}\text{s}\cdot\text{i}_3\left(\text{s}\right)=$$ $$\lim_{\text{s}\to0}\frac{\hat{\text{u}}_\text{i}\text{L}_1}{\text{s}\text{L}_1\text{L}_2+\text{L}_1\text{R}_1+\text{L}_2\text{R}_1}=\frac{\hat{\text{u}}_\text{i}}{\text{R}_1}\cdot\frac{\text{L}_1}{\text{L}_1+\text{L}_2}\tag{11}$$

$$\lim_{t\to\infty}\text{I}_3\left(t\right)=\frac{20}{5}\cdot\frac{10\cdot10^{-3}}{10\cdot10^{-3}+5\cdot10^{-3}}=\frac{8}{3}\approx2.66667\space\text{A}\tag{12}$$
• @Aaron no, take a look at $(9)$. Jan 30 at 20:48