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Find the current through the \$\mathrm{5 \space mH}\$ inductor when the circuit reaches a steady state

circuit

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?

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  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Arsenal
    Jun 5, 2020 at 7:36
  • \$\begingroup\$ Sometimes text books are wrong. \$\endgroup\$
    – Aaron
    Jan 29 at 19:57

2 Answers 2

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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.

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

schematic

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:

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

Using your values we get a steady state current of:

$$\lim_{t\to\infty}\text{I}_3\left(t\right)=\frac{20}{5}\cdot\frac{10\cdot10^{-3}}{10\cdot10^{-3}+5\cdot10^{-3}}=\frac{8}{3}\approx2.66667\space\text{A}\tag{12}$$

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  • \$\begingroup\$ Thanks for the downvote, maybe explaining it can be helpfull! \$\endgroup\$ Jun 6, 2020 at 16:11
  • \$\begingroup\$ I didn't down vote, but wouldn't using Laplace imply an AC source? \$\endgroup\$
    – Aaron
    Jan 29 at 18:44
  • \$\begingroup\$ @Aaron no, take a look at \$(9)\$. \$\endgroup\$ Jan 30 at 20:48

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