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Capacitors connected to batteries

  1. Oct 20, 2014 #1
    I want to know if my understanding is OK.

    If a capacitor is connected to a battery, the voltage across the capacitor will equal the battery's voltage, but this doesn't mean the capacitance is changing. The capacitance is a function of the capacitor alone (ignoring dielectrics). Right?

    If I have a capacitor connected to a battery and I double the distance between the plates then the energy will be halved, right? Even though the voltage has to equal the battery's.

    If I have a simple square circuit. And my capacitor comes first before my resistor (with respect to the positive plate on the battery), this shouldn't worry me, because the order wouldn't matter, right? Like, current is still in one direction only? And I could solve kirchoff's differential equation to find how my capacitor charges up?

    Like ##q=C\varepsilon-C\varepsilon e^{-t/RC}## ?

    And if I wanted to find the voltage across my capacitor as a function of ##t## I could divide that equation by ##C## right?

    Like: ##V=\varepsilon-\varepsilon e^{-t/RC}##

  2. jcsd
  3. Oct 20, 2014 #2


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    Staff: Mentor

    I haven't looked carefully at your equations, but you're right on the theory so if you've solved the differential equation properly you're in good shape.
  4. Oct 20, 2014 #3


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    Staff: Mentor

    That's right. Capacitance is independent of voltage.

    That's also correct. The order of the components don't matter here.
  5. Oct 20, 2014 #4
    I just came up with a new conceptual problem right now. The thing is I got a negative current when solving Kirchoff's voltage law, but it makes sense in this case as I just have to invert the direction of the current.

    My problem is this:

    I can start anywhere in my circuit to find currents on loops and build my system of equations from there on so long as I'm consistent with my directions right? And at the end I'll just have to interpret the negative/plus signs? right?

  6. Oct 22, 2014 #5


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    Gold Member

    Correct. A very important first step is to add arrows to your circuit diagram to define what you mean by +ve current and voltage. It doesn't matter if you are "wrong". As you said, it just means you may get a few -ve answers which you have to interpret.

    If you get some -ve values it might be tempting to go and change the arrows on your circuit diagram BUT that would be a mistake as it would make your previous working out wrong and could confuse an examiner.
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