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I am currently learning control system design. I have a lot of confusion about the initial conditions of the inductor in the circuit. Could someone please correctly solve this by getting the differential equations and then take the Laplace transform.

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  • \$\begingroup\$ You must show your attempt at constructing the differential equation. For the initial condition, assume the circuit has been connected for a long time, therefore the initial current is constant. \$\endgroup\$
    – Chu
    Apr 29 '19 at 6:28
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Using Faraday's law of induction, we can write:

$$-\text{V}\left(t\right)+\text{V}_\text{d}\left(t\right)+\text{R}\cdot\text{I}\left(t\right)=-\text{L}\cdot\text{I}'\left(t\right)\tag1$$

Where \$\text{I}\left(t\right)\$ is the input current (which is the same trough all the components because it is a series circuit).

Using Laplace transform we can write:

$$-\text{v}\left(\text{s}\right)+\text{v}_\text{d}\left(\text{s}\right)+\text{R}\cdot\text{i}\left(\text{s}\right)=-\text{L}\cdot\left(\text{s}\cdot\text{i}\left(\text{s}\right)-\text{I}\left(0\right)\right)\tag2$$

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  • \$\begingroup\$ Can you explain to me the significance of the I(0), the initial current for the inductor ? Because in some other examples I've seen they don't include that even though the Laplace of the derivative has it. \$\endgroup\$ Apr 29 '19 at 7:44
  • \$\begingroup\$ @MaizHalyym Well that is the initial condition in this circuit. So when there is an initial current trough the inductor it has to be placed there. \$\endgroup\$
    – Jan
    Apr 29 '19 at 8:00

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