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I understand that the current through a capacitor is given by the capacitance multiplied by the rate of change of voltage.

I also understand that this leads to infinite impedance for DC and virtually no impedance for higher frequencies.

What I haven't been able to find out is how this formula is derived from first principals.

Would someone be able to show me this?

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  • \$\begingroup\$ Please clarify what you mean for any capacitor \$ I = C \cdot \dfrac{\text{d}V}{\text{d}t} \$. You can't get more first principles. \$\endgroup\$ Feb 13 '19 at 10:51
  • \$\begingroup\$ @WarrenHill That's a fair point. I more mean how do we get to this equation? For instance, Ohm's law states that voltage is proportional to current and then we worked out that the constant of proportionality is resistance. I don't understand how this current equation can be derived in a similar way, I've only seen it flat out stated. \$\endgroup\$
    – user159015
    Feb 13 '19 at 11:17
  • \$\begingroup\$ Ask it in the Physics site. They should have known how that Q=VC relation came. \$\endgroup\$
    – Sadat Rafi
    Jun 23 '20 at 14:13
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Not sure if this is rigorous enough but

$$Q = CV = \int I \text{ d}t$$

Now differentiate with respect to time.

$$I = C \cdot \dfrac{\text{d}V}{\text{d}t}$$

If we assume \$C\$ is constant.

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\$Q=CV\$, and \$Q=It\$. Rearrange to get \$I=C*(V/t)\$

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  • \$\begingroup\$ Correct me if I'm wrong but that is a solid state equation not a differential equation. Can you just convert from V/T to dV/dT? \$\endgroup\$
    – user159015
    Feb 13 '19 at 11:15
  • \$\begingroup\$ Current is movement of charges. Ampere is a measurement of how many coloumbs charge that passes a point pr unit time - I = dQ/dt The magnitude of charge Q on a capacitor is proportional to Capacitance and voltage across the plates. Q = CV. dQ/dt = C dV/dt. Subsitute dQ/dt and get I = C dV/dt \$\endgroup\$
    – MrGerber
    Feb 13 '19 at 11:59
  • \$\begingroup\$ Read more about charge definition on en.wikipedia.org/wiki/… and capacitance definition on en.wikipedia.org/wiki/Capacitance \$\endgroup\$
    – MrGerber
    Feb 13 '19 at 12:04

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