0
\$\begingroup\$

I am a lot confused about how to assign the current of a capacitor and the sign of its constitutive relation, because usually in physics and in engineering there are different conventions. There are some topics here about it but they usually refer to one of these conventions, and I want to understand the link between them and how to apply them correctly.

Precisely, in engineering I have been always told that this convention will work (since it is the passive sign convention, and a capacitor is a passive device):

Voltage across passive components decreases in the direction of current

enter image description here

So, if we define current and voltage with this polarity (as in picture), we may write that:

enter image description here

It is a very good and practical convention: we know how to choose current and voltage polarities, and their relationship is positive, and there is no ambiguity of sign. I have used it a lot in the analysis of many complex circuits, with also other elements (diodes, transistors etc).

It seems quite clear. But now, let's consider these two typical situations, in which the previous convention is applied:

1) Capacitor charging:

enter image description here

Suppose the capacitor discharged at t = 0, in which the switch is closed. It will start charging and its voltage may be found with the following analysis:

KVL:

enter image description here

Let's put in it the relation:

enter image description here

So:

enter image description here

and finally:

enter image description here

Well, it works.

2) Capacitor discharging: let's apply the previous convention to this new situation, in which the circuit is closed when the capacitor has a certain voltage V0

enter image description here

enter image description here

Let's put in it the relation:

enter image description here

So:

enter image description here

it works: the current is negative, and it is correct because it physically flows in the opposite direction since the capacitor is discharging.

In physics I have seen a different analysis: the capacitor is discharging, and so its constitutive relation will be:

enter image description here

So:

enter image description here

Obviously it does not work. It will work if I take the current in the opposite direction.

So, my final questions are:

1) Which is the link between these two conventions?

2) How can I apply the "physics" convention without sign mistakes?

3) Why does the first convention work also in case the capacitor is discharging, which means that it is working as an active device?

4) How can apply the physics convention in a general circuit (with other components) in which a capacitor is charging at some times and discharging at other times?

5) Obviously I have all these doubts for an inductor too...

About the "physicists" convention, I have found for instance this material (see page 2). It says, about the discharge of a capacitor:

enter image description here

\$\endgroup\$
7
  • \$\begingroup\$ See if this helps: electronics.stackexchange.com/questions/253433/… \$\endgroup\$ – Big6 Apr 18 '20 at 16:09
  • 3
    \$\begingroup\$ You should share exactly where you saw this "physics convention" used. The context of how they explained whatever problem they were presenting might reveal why they defined their variables the way they did. \$\endgroup\$ – The Photon Apr 18 '20 at 16:13
  • 1
    \$\begingroup\$ Try this also: electronics.stackexchange.com/questions/281549/… \$\endgroup\$ – Big6 Apr 18 '20 at 16:15
  • \$\begingroup\$ It sounds like you should really be asking the physicists about this, if it is their problem. For question 3, when a capacitor is discharging it is supplying power to the rest of the circuit but that does not make it an active device. \$\endgroup\$ – Elliot Alderson Apr 18 '20 at 17:18
  • \$\begingroup\$ When you charge a capacitor, it behaves as a passive element (power consumer) that "steals" power (it accumullates but still "steals" it in contrast to the resistor that dissipates power). The current enters the capacitor positive plate. When you discharge a capacitor, it behaves as a source... an active element (in the sense that it produces power). Now the current exits the capacitor positive plate. \$\endgroup\$ – Circuit fantasist Apr 18 '20 at 20:42
-2
\$\begingroup\$

I hope this is the correct answer that you need:

There are 2 types of capacitors, polarized and not polarized.

The polarized capacitor has its signs on it. If you switch them and connect the capacitor - to the wire +, then the capacitor will get charged with negative voltage compared to ground.

The not polarized capacitor charges depending on which end is connected to the + of the power supply. The + end will be positive compared to ground.

This of course is accepting the capacitor is in paralel to the power supply + and ground.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Welcome to EE.SE, but I don't think you have understood the question. Also your explanation assumes that power supply negative is always the ground which is not always the case. \$\endgroup\$ – Transistor Apr 18 '20 at 20:10
  • \$\begingroup\$ I have noted this: This of course is accepting the capacitor is in paralel to the power supply + and ground. \$\endgroup\$ – CFCBazar com Apr 19 '20 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.