Capacitors connected to batteries

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    Batteries Capacitors
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Discussion Overview

The discussion revolves around the behavior of capacitors when connected to batteries, focusing on concepts such as capacitance, voltage, current direction, and Kirchhoff's laws. Participants explore theoretical aspects, mathematical equations, and circuit analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that the voltage across a capacitor connected to a battery equals the battery's voltage, and that capacitance remains unchanged regardless of voltage.
  • Another participant agrees with the assertion about capacitance being independent of voltage.
  • A participant questions whether the order of components in a circuit affects current direction and concludes that it does not, as long as the current direction is consistently defined.
  • One participant describes encountering a negative current when applying Kirchhoff's voltage law and seeks clarification on interpreting negative signs in circuit analysis.
  • Another participant confirms that defining current and voltage directions with arrows in circuit diagrams is crucial, and emphasizes that negative values should be interpreted rather than changing the diagram's direction arrows.

Areas of Agreement / Disagreement

Participants generally agree on the independence of capacitance from voltage and the irrelevance of component order in determining current direction. However, there is an ongoing exploration of how to interpret negative values in circuit analysis, indicating some uncertainty in that area.

Contextual Notes

Participants discuss the implications of negative current values and the importance of consistent direction definitions in circuit analysis, but do not resolve the nuances of these interpretations.

davidbenari
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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 kirchhoffs 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}##

Thanks.
 
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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.
 
davidbenari said:
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?

That's right. Capacitance is independent of voltage.

davidbenari said:
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 kirchhoffs differential equation to find how my capacitor charges up?

That's also correct. The order of the components don't matter here.
 
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?

Thanks.
 
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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