So I was trying to derive the exponential decay equation for a discharging capacitor and realised that I would only get the correct answer if I used a negative current, that is to say the direction of the current opposes the direction of the voltage applied by the capacitor?(this is probably where the problem is). Here is the equation: $$Vc(t) - (-RC\frac{dVc(t)}{dt}) = 0$$

I have also visited links to similar questions and saw that the negative current means that the discharging current is opposite to the charging current. But what if I start with a charged capacitor? In this case am I not free to define the direction of the current in which ever way I want? In short, I would like to clarify whether there is a potential gain or a potential drop at each of the elements(a capacitor and a resistor).

To elaborate:If I have a circuit with only a charged capacitor that is discharging and a resistor, and I perform KVL around the loop in the direction of the actual current, following passive sign convention. Do I not end up with the equation:

$$Vc(t) - (RC\frac{dVc(t)}{dt}) = 0$$ why is this equation not valid?

  • \$\begingroup\$ The current isn't based on the charge on the capacitor, but on what you're doing to it. \$\endgroup\$ Commented Jan 21, 2017 at 2:18

4 Answers 4


I once had the same doubt, but in short, it has to do with the passive sign convention.

This is the circuit that you have:


simulate this circuit – Schematic created using CircuitLab

See that instead of using KVL, I am using KCL for now. I defined the node \$v_o\$. I have defined my currents in the direction shown, but you can certainly choose other directions. It follows that:

$$ i_c + i_R = 0$$

And you could now plug in what \$i_c\$ and \$i_R\$ are, to get


And that's the differential equation that will give you the well known solution for a discharging capacitor.

Why does it work out for KCL and you can't seem to get it to work using KVL?

The trick is in the use of the positive sign convention. Passive devices have a positive current and voltage relationship when the 'current is going into the positive terminal and comes out of the negative terminal'

Since the current is going into the elements through the + terminal and comes out through the negative terminal then the current is positive, by the PSC.

Here is an excerpt from Nilsson-Riedel Electric Circuits book

enter image description here

So when you see the capacitor has the \$i_c=+C\dfrac{dv}{dt}\$ (notice the +), that's the definition following the passive sign convention where the current enters the positive terminal.

If you were to use KVL, take a look a the following approach:


simulate this circuit

Which is the same as the first one I drew but I added another current defintion, the one I will use for the KVL loop. I named that current \$i_s\$ (in red), and let's do KVL:

$$ v_o-i_sR=0$$

That's where you get confused. Now, you can see from the circuit that \$i_s\$ goes in the opposite direction compared to the \$i_c\$ current (definition by psc), that is,

\$i_c=-i_s\$ or \$-i_c=i_s\$

And since the current \$i_s\$ is in the same direction as \$i_r\$,

$$i_s=i_r $$

And if you plug the \$i_s\$ and \$i_c\$ relationship, you end up with the right differential equation:

$$ v_o-(-i_c)R=0$$ $$ v_o+i_cR=0$$ $$C\dfrac{dv_o(t)}{dt}+\dfrac{v_o}{R}=0$$

Hope it helps.

  • \$\begingroup\$ I am still unsure about ic, wouldn't a discharging capacitor act like a voltage source with current flowing from + to -? \$\endgroup\$
    – Frank
    Commented Jan 21, 2017 at 8:15
  • 2
    \$\begingroup\$ @Frank, for a capacitor, \$I=\frac{1}{C}\frac{\rm{d}V}{\rm{d}t}\$. The direction of the current relates to whether the voltage is increasing or decreasing, not to whether the voltage is positive or negative. \$\endgroup\$
    – The Photon
    Commented Jan 21, 2017 at 8:38
  • \$\begingroup\$ @Frank Yeah, it acts as a voltage source. The current flowing out of it (out of the + term), is what I named \$i_s\$. But that current goes opposite direction compared to the definition of the passive sign convention (what I named \$i_c\$. Why do you think, a voltage source delivering power and following positive sign convention, has negative power? Because the current it's coming out of the positive terminal, then Ps=-I*V. If current were going into the + terminal, then the source absorbs power. \$\endgroup\$
    – Big6
    Commented Jan 21, 2017 at 15:07

A drawing of positive directions is needed. Let's assume that a capcitor has a positive voltage between its poles. Be the positive current charging or discharging, it's defined in that drawing. Charging in everyday talk has no unique current direction. Charging in everyday talk is the situation where the voltage between capacitor poles drifts further from zero.

Stick with these when U is positive

enter image description here

For more complex circuit the drawing is still needed. You did not provide one!No possiblity to judge which is the right equation.

Choose one of my drawings, add a resistor and write Ohm's law for resistor current. Write that current to be equal with my I. Be sure that you have written Ohm's law using the same positive U and I directions than stands in my drawing that you selected. Then you get the right equation for the selected drawing.

Let's add a resistor to the leftmost drawing. There Ohm's law is: I=-U/R because the positive current goes into the resistor at that end where is + voltage compared to the other end.

The selected drawing leads to your equation. The most often seen equation is true when a resistor is added to my rightmost drawing.


No, you don't end up with that equation.

If you reverse the orientation of your "probes" on the capacitor, such that you see negative current instead of positive, you'll also see negative voltage instead of positive.

That is, every appearance of \$Vc(t)\$ will change its sign, resulting in an equation which is exactly equivalent to the first.


I recently had the urge to go back and understand the raw basics of where the capacitor/resistor charge and discharge equations came from. After a quick look online, it was easy to find and understand the simple circuit of a capacitor charging from a fixed DC voltage through a resistor. Not so easy was finding much on the discharge equation derivation.

Finally it dawned on me that math modeling must be done in a way that reflects the actual original. So, my (successful) approach was thus:

[![enter image description here][1]][1]

Simple circuit consisting of a capacitor "C" starting off with a charge of Vx, is in parallel with a resistor "R". The voltage on the capacitor (at any point in time) is denoted by Vc. Therefore, $$ Vc = Vr $$ And Ohm's law dictates that $$ Vr = (R)(Ir) $$.

Now here's where it all made perfect sense to me: Starting with
$$ Ir = Ic $$ and $$ Ic = C \left(\frac{\Delta(Vc)}{\Delta(t)}\right) $$ It's very important to realize that since the capacitor is discharging, its voltage, Vc, is decreasing in time. And the magic silver bullet is the fact that because Vc decreases as time "t" increases, \$ \frac{\Delta(Vc)}{\Delta(t)} \$ obviously must be a negative value. No PSC (passive sign convention) needed; nor ASC (Active Sign Convention) needed whatsoever.

$$ Vc = RC\frac {-\Delta(Vc)}{\Delta(t)} $$

After going through the same motions (mathematically speaking) one goes through to obtain the capacitor charging equations, the correct discharge equation is obtained.

The above may not be a "proper" way to explain/analyze the discharge circuit, but it does work, and it works for me. Perhaps you'll find it acceptable as well.

For the benefit of those who want to see the rest of the omitted steps above, here they are:

$$ Vc = Vr $$ $$ Vc = (R)(Ir) $$ $$ Ir = Ic \quad and \quad Ic=C\frac {-\Delta(Vc)}{\Delta(t)} $$ $$ Therefore, \quad Vc = RC\frac {-\Delta(Vc)}{\Delta(t)} $$ $$ \Delta(t) = \frac{RC}{Vc}(-\Delta(Vc)) $$ $$ \int\Delta(t) = \int\frac{-RC}{Vc}\Delta(Vc) $$ $$ \int\Delta(t) = (-RC)\int\frac{1}{Vc}\Delta(Vc) $$ $$ t = (-RC)(ln(Vc))\;+ K$$ $$ At\quad t = 0,\quad K=(RC)(ln(Vx))$$ $$ Therefore,\quad t = (-RC)(ln(Vc))+(RC)(ln(Vx))$$ $$ \frac{t}{RC} = (ln(Vx))-(ln(Vc))$$ $$ e^\frac{t}{RC} = \frac{Vx}{Vc}$$ And finally, $$ Vc=(Vx)(e^\frac{-t}{RC})$$

To answer the original question, which was "why is this equation not a valid starting point for deriving the equation of a capacitor discharging through a resistor?" --->

The equation $$ Vc(t) - \left(RC\frac{dVc(t)}{dt}\right) = 0 $$ can be rewritten as $$ Vc(t) = \left(RC\frac{dVc(t)}{dt}\right)$$ As such, \$ Vc(t) \$ will INCREASE in time because the term \$ \frac {dVc(t)}{dt} \$ is positive. Since \$ Vc(t)\$ actually DECREASES in time, that's why the starting equation is invalid.



  [1]: https://i.sstatic.net/S7fQJ.png
  • \$\begingroup\$ Welcome to EE.SE. If you're going to hang around you might as well learn some MathJAX for your equations. You'll love it. \$\endgroup\$
    – Transistor
    Commented Sep 22, 2017 at 20:36
  • \$\begingroup\$ Transistor, Thanx for the MathJAX suggestion. I spent a few hours of quality time trying my best to figure out how to get it to work. On the Mathjax.org site, there's talk of installing it "on your server" and opening files up "in your webpage" -- I don't have a server, nor a webpage, nor any experience with either. Any suggestions on how I can proceed and possibly make use of this wonderful tool? \$\endgroup\$
    – Eugeneus
    Commented Sep 26, 2017 at 5:37
  • \$\begingroup\$ Did you miss the link in my comment? For this site you just put your equations between $$ tags or, for inline, between \$ tags. e.g. \$ V_C = R C \frac {-dV_C}{dt} \$gives \$ V_C = R C \frac {-dV_C}{dt} \$. You can also see how others have done it by hitting the "edit" links on their posts. \$\endgroup\$
    – Transistor
    Commented Sep 26, 2017 at 6:45

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