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There is something I can't grasp regarding capacitors. Say for example I'm using a simple rc circuit that has one 5v dc power source, a resistor, and a capacitor. The capacitor will charge until it reaches 5v, then cut all current. What I don't understand is how the capacitor can receive the charge in the first place? Isn't a capacitor effectively an open circuit, therefore there shouldn't be any flow and the capacitor shouldn't be able to charge in the first place? Isn't it analogous to a switch that is in the open state in a circuit?

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    \$\begingroup\$ If the capacitor is charging, then by definition you have a time-varying voltage in your circuit, which is not DC. Connecting the battery is equivalent to applying a step function. Again, not DC. \$\endgroup\$
    – Adam Haun
    Commented Sep 27, 2018 at 20:53
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    \$\begingroup\$ Current doesn't flow through the capacitor - the dielectric is an insulator. Charge flows onto the plates. As the charge builds up, so does the voltage across the capacitor, and the direct current reduces since the voltage across the series resistor decreases; falling to zero when the capacitor is fully charged. \$\endgroup\$
    – Chu
    Commented Sep 27, 2018 at 23:49

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Imagine two metal plates very close together, but not touching. They are so close that the electrons on one plate can 'feel' electrons on the other.

When you apply a voltage, you are pushing electrons onto one plate, which repel the electrons on the other plate. You can imagine that for every electron you push onto one plate, an electron leaves the other.

This happens until the voltage is no longer sufficient to push anymore electrons onto the plate. This happens because you have pushed so many electrons onto that plate that the force of them repelling each other is equal to the voltage you have applied.

In a DC circuit, the charged capacitor is similar to an open switch as no more current will flow.

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    \$\begingroup\$ Good answer. You should add in the capacitor formula relating C, Area and distance. \$\endgroup\$
    – Transistor
    Commented Sep 27, 2018 at 22:18
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    \$\begingroup\$ Superb and very clear answer. \$\endgroup\$ Commented Sep 28, 2018 at 3:55
  • \$\begingroup\$ This is the best answer I've found. Or I've read so many that this one solidified it. \$\endgroup\$
    – johnny
    Commented Oct 5, 2018 at 14:32

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