This may be a very basic question but my misunderstanding is due to the difference between how physics explains the charging of capacitors and how we analyze circuits in circuit analysis.

According to physics, when a battery is connected to an uncharged capacitor, electrons flow from one plate to the other creating an electric field, and hence a potential difference across the capacitor. This potential difference starts at 0 and increases until it reaches the voltage of the battery.

However, this means that for the time the capacitor is charging, the potential difference across it is different from the potential difference of the battery. According to circuit analysis (or at least the lumped-element model) this connection is not valid. The definition of 2 elements being parallel is that they have the same voltage.

The only solution I can think of right now is that it is indeed valid to have this connection however in real-life circuits there is no ideal short circuits so if we connect a battery to a capacitor, the wires have a bit of resistance and so we can model them as resistors in the circuit. So when the battery is connected, the resistance will have all the battery's voltage at first while the capacitor will have 0 volts. The current that flows is due to the capacitor charging (and will be very high because the resistance of the wire is very low). As the capacitor is charging, its voltage increases while the voltage of the resistor decreases until the capacitor takes on all the battery's voltage causing the current to be zero and so the resistor's voltage becomes 0.

And that's why analyzing the step response of a capacitor, there's always a resistor (usually with a high resistance, on the order of 10 kilo Ohms) connected in series with the capacitor.

Is this really the case? Or am I missing something?

  • \$\begingroup\$ Can you take a glance at some of the prior discussion on related topics such as this and let us know what additional doubts you have? \$\endgroup\$
    – nanofarad
    Sep 2 '20 at 18:54
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    \$\begingroup\$ The capacitor will have some series resistance to save your day. \$\endgroup\$
    – winny
    Sep 2 '20 at 18:58
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    \$\begingroup\$ The battery won't be ideal voltage source either. So in real life, there will be some resistance in the system (in battery, wires and capacitor) which defines the peak current of the step response, without resistance the peak current would be undefined (dividing voltage with zero resistance). \$\endgroup\$
    – Justme
    Sep 2 '20 at 19:03

Good question. If you connect an IDEAL voltage source to an IDEAL capacitor you get exactly the situation that you describe. The IDEAL answer is that there will be an INFINITE amount of current for an INFINITELY short amount of time to get it all sorted out.

In practice that doesn't happen since IDEAL voltage sources and capacitors do not exist. All real voltage sources have some finite (non-zero) source impedance which limits the current and results in normal exponential charging of the capacitor. The smaller the source impedance the quicker the capacitor will charge and the higher the initial charging current will be.

And there is always option B: something is just going to blow up. Connecting a large uncharged capacitor to a beefy power source with a very low source impedance is best observed from a safe distance until the fireworks have subsided. In other words: DON'T DO THIS.


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