Please consider the following RC circuit as context:

enter image description here Assume that the circuit has been connected for a long time. If switch S has been opened at $t=0,$ the differential equation used to solve for the charge on the capacitor $Q$ would be, by using Kirchhoff's loop rule: $$\frac{Q}{C}-R\frac{dQ}{dt}=0, \quad Q(0)=C\mathcal{E}$$

since, discarding the portion to the left of the capacitor, the voltage drop through the capacitor would oppose that of the resistor following the clockwise current flow. However, a professor told me that the right differential equation in this case would be: $$\frac{Q}{C}+R\frac{dQ}{dt}=0, \quad Q(0)=C\mathcal{E}$$ I simply do not understand how the voltage drop of the capacitor is negative from its bottom to top plate. Wouldn't it act the same as a battery? Namely, wouldn't it add the the voltage to the circuit?

  • 1
    $\begingroup$ The problem here is that the reference polarities for the capacitor and resistor voltage variables are not marked on the schematic. In addition, the capacitor current $i_C = dQ/dt$ is the current downwards through the capacitor if $Q$ is the charge on the top-most plate of the capacitor. KVL then gives the equation your professor said was correct. I've detailed all of this in an answer. $\endgroup$ May 31, 2020 at 22:08

4 Answers 4


I believe that your Professor is correct. The equation by bobD is correct but what you are missing is that the charge on capacitor on that time is Q and therefore current flowing through that instant is $d(Q(0)-Q)/dt$ and that gives your correct equation. Even if you consider your equation correct then when you integrate it the charge will increase exponentially which is not possible. HOPE THIS HELPS

  • $\begingroup$ Note that answers can be edited so the statement "The equation by bobD is correct" might be right, wrong, or nonsense if the answer by bobD is edited after you wrote that. $\endgroup$ May 31, 2020 at 22:09
  • $\begingroup$ @Alfred Centauri Thank you I will take care of it in next quetions $\endgroup$ Jun 1, 2020 at 6:27

At time t=0 the voltage polarity of the capacitor is the same as that of the battery. So the battery discharges with current flowing clockwise through the resistor. Per Kirchhoff's voltage law $+V_{C}(t)-i(t)R=0$. But $i(t)=dQ/dt$, and $dQ/dt$ is negative (current is decreasing), which makes the second term in the differential equation positive. So your prof is correct.

Hope this helps.

  • $\begingroup$ I would have thought so, but solving the differential equation that I set up would yield the wrong discharge equation, where the exponent is positive instead of negative. How would you conceptualize the negative voltage drop across the capacitor? $\endgroup$ May 31, 2020 at 16:28
  • $\begingroup$ Actually, my equation as written is correct but I wasn't looking at the differential equation which was correct. I will update. $\endgroup$
    – Bob D
    May 31, 2020 at 16:33
  • $\begingroup$ Should the second sentence be "So the capacitor discharges..."? Also, I don't think the reasoning in your 4th sentence is sound. If $i(t)$ is the current clockwise, and if $Q$ is the charge on the top plate (consistent with your KVL equation), then $i(t) = -dQ/dt$ $\endgroup$ May 31, 2020 at 21:59

The schematic is missing something quite important - the reference polarity for the circuit element voltages.

If you do this, there will be no question of which equation is correct.

For example, denote the voltage across the capacitor as $v_C$ and place a "+" symbol at the top-most terminal of the capacitor. This tells us is that $v_C$ is positive when the potential at the top-most terminal is the more positive.

Put another way, the "+" symbol indicates that if the "red" lead of the voltmeter is connected there (and the "black" lead is connected to the other terminal), the voltmeter will read $v_C$. Note that if you reverse the connection, the voltmeter reads $-v_C$.

Similarly, denote the voltage across the resistor as $v_R$ and place a "+" symbol at the top-most terminal of the resistor.

(Before you read further, make you've marked up your schematic so you can reference it as you read the rest.)

Now, with these reference polarities, KVL counter-clockwise around the capacitor-resistor loop (starting at the top) yields

$$v_C - v_R = 0$$


$$v_C = \frac{Q}{C}$$

where $Q$ is the charge on the top-most plate (this matches the "+" sign on the top-most terminal) and

$$v_R = i_RR$$

where $i_R$ is the current into the top-most terminal (this matches the "+" sign on the top-most terminal by the passive sign convention).

Recall that

$$i_C = \frac{dQ}{dt}$$

and (important!) this is the current into the top-most terminal of the capacitor (again, by the passive sign convention). Thus, by KCL,

$$i_R = - i_C = -\frac{dQ}{dt}$$

and it follows that the correct equation is

$$\frac{Q}{C} + R\frac{dQ}{dt} = 0$$

with solution

$$Q(t) = C\mathcal{E}e^{-t/RC},\qquad t\ge 0$$


I believe your professor is right. The solution of his differential equation would be a damped exponential $$Q(t)=Q(0)e^{-t/RC}$$ which makes senses as a discharging capacitor.

But the solution of your differential equation would be a growing exponential $$Q(t)=Q(0)e^{+t/RC}$$ which means the capacitor's charge would grow infinitely. This cannot be correct.


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