How would I calculate the electro-static potential for this simple circuit during the transient (during the time the capacitor charges)? Do I just use Ohm's law?


simulate this circuit – Schematic created using CircuitLab

My thought process was as follows:

  • We know when the capacitor is charged it will have a "charge" of 1V * 1μF (Psi aka. Q)
  • So we have a current (I) of the value Psi / Δt during the transient
  • I = e / Re = I * R (e is the potential, aka. voltage, of the capacitor and R is the resistance of the conductor)

I wrote down my full thought-process here: http://circuits.icidasset.com/circuits/01-basics/002-capacitor

Does this seem correct, or is there a better way?

  • 2
    \$\begingroup\$ As drawn you have only 'ideal' components. You forgot to add the inductance the circuit certainly has. The initial charge on the capacitor is zero, not 1V*1uF (*10^-6). The initial voltage is zero. Adding a step voltage of 1V from a perfect source into this ideal circuit with no resistance but finite inductance due to the loop nature of the circuit (not shown on your diagram) would produce oscillation and bring about the destruction of the Universe. Add a small series resistance (say) 1 ohm and this will bring the circuit back into reality. Then you can use the conventional equations. \$\endgroup\$ Dec 22, 2016 at 11:28
  • \$\begingroup\$ Where is R in your circuit and what is the "electro-static potential" you are trying to calculate? \$\endgroup\$
    – Andy aka
    Dec 22, 2016 at 11:33
  • \$\begingroup\$ @JImDearden Thanks. Yeah I meant the value when the capacitor charged (the condition after the charge). And the conductor has resistance and inductance, has it not? R in my equations would be the resistance of the conductor. \$\endgroup\$
    – Icid
    Dec 22, 2016 at 11:34
  • \$\begingroup\$ @Andyaka Updated the question, R is the resistance of the conductor. And the electro-static potential is the voltage of the capacitor during the transient. \$\endgroup\$
    – Icid
    Dec 22, 2016 at 11:36
  • 1
    \$\begingroup\$ You need to show the resistance in the circuit diagram. There are three sources of resistance. The voltage source (because it can't supply infinite current), the connecting wire (because we aren't using a superconductor) and the internal series resistance (ESR) of the capacitor. These can be lumped together and shown as one resistor. Also, a capacitor never reaches the final value of the voltage. We assume its fully charged after 5 time constants (>99%) of final voltage. For a time constant you need a value for R. \$\endgroup\$ Dec 22, 2016 at 11:44

1 Answer 1


The capacitor charges like this: -

enter image description here

The exact formula is: -

\$V_C = V_S(1-e^{-t/RC})\$

  • \$\begingroup\$ Thanks! Could you clarify what the last part of that formula (e...) exactly is? \$\endgroup\$
    – Icid
    Dec 22, 2016 at 11:51
  • 1
    \$\begingroup\$ "e" is 2.71828 i.e. the natural exponential number (en.wikipedia.org/wiki/E_(mathematical_constant)). "t" is time. \$\endgroup\$
    – Andy aka
    Dec 22, 2016 at 11:53

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