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I am in the process of solving a DE for a circuit.

So: the circuit starts with a voltage source of 5 VDC and is allowed to reach steady state. At time t = 0, the source is switched out for a 10 VDC source. schematic

I am in the process of finding the initial conditions: $$i_L(0^+)\ \ \mathrm{and}\ \ i_L'(0^+)$$

Clearly, $$v_C(0^-) = 0\ \mathrm{V}\ \ \mathrm{and}\ \ i_L(0^-) = \frac{5\ \mathrm{V}}{100 \ \Omega}$$

So $$i_L(0^+) = 50\ \mathrm{mA}$$

Now, to find iL', I said $$v_L = Li_L'\ \ \mathrm{and}\ \ v_L = v_C$$

So $$i_L'(0^+)=\frac{v_C(0^-)}{L} = 0\ \frac{\mathrm{A}}{\mathrm{s}}$$

But, if you simulate this circuit and flip the switch, the slope of the current through the inductor is clearly not zero.

Am I doing something wrong, or is there something subtle happening?

Thanks

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  • \$\begingroup\$ The current in an inductor can't change instantly, but the derivative can. \$\endgroup\$ – Greg d'Eon Jan 19 '16 at 1:00
  • \$\begingroup\$ Yes, I am aware of that. The problem I'm having is that the (apparently) correct math shows an initial derivative of zero, even though the simulator disagrees. \$\endgroup\$ – Mahkoe Jan 19 '16 at 1:03
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[ last line should be i'L(0+) ] Your calculations are correct. At t=0+, there is 50 mA in the inductor, but 0 V across it (and the capacitor). So i'(L) =0. Now, the voltage across the capacitor will increase (because supply increased from 5 to 10 V; the 'extra' 50 mA will flow in the capacitor. The cap voltage will rise, and then the inductor current will increase.

In a small region around 0+, the cap V will increase linearly, and inductor current quadratically with time, however overall the result will be (exponentially decaying) sinusoids.

Clearly at the extreme of a sinusoid, the derivative is 0, yet since the 2nd derivative isn't 0, the derivative will change, and so will the value.

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Are you sure you did the simulation correctly? I found that at the moment the voltage source changes, the slope of the inductor current is zero:

CircuitLab sim

You can see that the oscillation starts in the trough of a sine wave, where the derivative is zero. Here's a longer view showing the exponential decay of the oscillation:

Another CircuitLab sim

UPDATE: I get the same result with Falstad:

Falstad simulation capture

It's harder to see because you can't zoom in easily. I had to increase the inductance and capacitance to get this much. But you can see that the ringing starts in the trough of a sine wave, where (as jp314 points out) the derivative is zero. If you crank up the plot speed and move the simulation speed around you can get a better view:

A zoomed in Falstad sim

Falstad is great for visualization, but getting precise results is awkward. For transient simulations, I recommend using CircuitLab, which you can use for free in an EE.SE answer box.

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  • \$\begingroup\$ Huh. I guess my simulation program doesn't have reliable results then. (I use the falstad circuit simulator: falstad.com/circuit/circuitjs.html) \$\endgroup\$ – Mahkoe Jan 20 '16 at 0:17
  • \$\begingroup\$ I updated my answer with Falstad results. \$\endgroup\$ – Adam Haun Jan 20 '16 at 3:09

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