Capacitors connected to batteries

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When a capacitor is connected to a battery, the voltage across the capacitor matches the battery's voltage, but the capacitance remains unchanged. Doubling the distance between the capacitor plates will halve the energy stored, despite the voltage being constant. The order of components in a circuit, such as a capacitor before a resistor, does not affect the current direction, and Kirchhoff's equations can be used to analyze the circuit's behavior. Negative current values obtained during calculations indicate the need to interpret current direction rather than changing circuit diagram arrows. Consistency in defining current and voltage directions is crucial for accurate analysis.
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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