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The formula for voltage on an inductor is stated as

v = L (inductance) x (di / dt)

and current for a capacitor is

i = C (capacitance) x (dv / dt)

But I don't understand what the (dv/dt) part of this equation should be.

For example

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit, if I wanted to know the current on the capacitor at some point in time, how would this be calculated. Would we count the "change in voltage" as going from 0v to the supply voltage in one unit of time? Or would it happen more slowly? How do we know how slowly?

What if there was a switch in this circuit, so there was 0v and then the voltage changed to 1v?

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1 Answer 1

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Your circuit diagram shows an ideal voltage source with 1 V output, from \$t=-\infty\$ to \$t=\infty\$. Since the voltage is constantly 1 V, \$\frac{dV}{dt}=0\$.

If you want a nonzero \$\frac{dV}{dt}\$ you should include some mechanism in the circuit to change the votlage, or use a source with a varying output voltage.

In early classes, this is often done with switches. In the real world it might also be done with transistors or various other mechanisms.

What if there was a switch in this circuit, so there was 0v and then the voltage changed to 1v?

Then the rate of change of the capacitor voltage would depend on parasitic resistances you haven't shown in the circuit. The voltage source has an effective resistance of some kind, the capacitor has an effective series resistance (ESR), and the wires have resistance.

If there's no discrete resistor added to the circuit, these unknown parasitics will determine the rate the capacitor charges at.

If these parasitic resitances are very low, then the parasitic inductance of the wires and of the capacitor could also play a role.

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  • \$\begingroup\$ I have edited my comment, I meant to include a switch in the schematic. \$\endgroup\$
    – user221241
    May 9, 2019 at 1:02
  • \$\begingroup\$ But in the "ideal" case it would be instant? \$\endgroup\$
    – user221241
    May 9, 2019 at 1:38
  • \$\begingroup\$ @user221241, yes, if the voltage source is able to deliver infinite current. There is a special case if the voltage source is actually another pre-charged capacitor where you can show that for conservation-of-energy reasons, half of the energy in the system must be absorbed by the parasitic resistance, no matter how small that resistance is. \$\endgroup\$
    – The Photon
    May 9, 2019 at 1:54
  • \$\begingroup\$ If there was a total resistance of 1ohm, how would we do the calculation? \$\endgroup\$
    – user221241
    May 9, 2019 at 1:56
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    \$\begingroup\$ allaboutcircuits.com/textbook/direct-current/chpt-16/… \$\endgroup\$
    – The Photon
    May 9, 2019 at 2:13

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