I have the next circuit:

enter image description here

The textbook says that the capacitor is initially charged, but the current directions was selected as in schematic, why? If capacitor is charged and "+" is on upper node then the capacitor must be a voltage source in this circuit and "iC" must go up to node.

The textbook solution is:

enter image description here

Here an author derivate the result equation. But stop at a second and look. He use hard-coded iC direction. What will be if I try to derivate result myself and select the iC to "up"? The KCL will be: $$ C*\frac {\partial v_{C}}{\partial t} = \frac {v}{R} $$.

Solving, get:

$$v(t) = A*e^{\frac{t}{R*C}} $$

And there is no "-" in "e" power. The other current direction give another tau. I know that one task may be solved with different solutions, but anyone can't take in mind not general solutions in mind for every task. I want to see where an error and how to changing directions of currents I can get proper results with method of that textbook.

For more clarification of my problem I redraw the schematic:

enter image description here

Now I think so: as the capacitor is charged and the external voltage source is turned off then I can think about capacitor as a voltage source with it's own stored charge and the "iC" current begin going through the circuit in one direction with "iR" and the capacitor is discharging through the resistor. The textbook recommends to write KVL with components sign equal to first achieved through loop, in my case it would be:

$$ -\frac{1}{C}*\int {i_C}{\partial t} + i_R*R = 0$$

solving it I get:

$$v(t) = A*e^{\frac{t}{R*C}} $$

Here ther is no "-" sign in exponent and that equation shows that the voltage on "R" will be increasing and stay on max. values. But it's incorrect! Where is an error? It seems to be I can't interpret a capacitor with stored voltage as an active component.

  • 1
    \$\begingroup\$ What if the \$I_c\$ arrow points down but the value of \$I_c\$ is negative? My point: I can define that \$I_c\$ flows counterclockwise while the actual current flows clockwise. The arrow points in the direction that the current would have if it was positive. No one says it actually has to be positive. \$\endgroup\$ Dec 19, 2017 at 9:21
  • \$\begingroup\$ @Bimpelrekkie I know that I can define any direction but the textbooks use that initial assumtions to derive transient response and if choose another direction the anwer will not only differ by sign but for tau too. Why the books always use any "right" directions that give right results? \$\endgroup\$
    – MaxMil
    Dec 19, 2017 at 9:28
  • \$\begingroup\$ The people writing books and giving classes are experienced so they already know the answer and what to do to get there more easily. In the end the direction of a current is irrelevant as long as you keep the sign (+ or -) correct. \$\endgroup\$ Dec 19, 2017 at 9:31
  • \$\begingroup\$ @Bimpelrekkie Look an update to question. What do you think? \$\endgroup\$
    – MaxMil
    Dec 19, 2017 at 13:05
  • \$\begingroup\$ Your question is too long for me to read the whole thing right now. But in a circuit with R and C only, Tau is always just R*C. If you remember that, it may save you some time in the future. \$\endgroup\$
    – mkeith
    Dec 21, 2017 at 8:35

4 Answers 4


Technically speaking, your textbook is probably using the convention typical for the loads, ie current enters from the + terminal and exits the - terminal. For the generators the opposite is true.

A capacitor is a load, although in this case it is releasing the energy stored in it and acting as a "generator". This is why your book is using that convention.

Note, however, that this doesn't really matter that much since you can easily:

  1. Realise that current direction in a circuit is purely a convention and, once the result is found, the sign of the current will indicate the actual direction of the "conventional current". As long as you remain consistent with the load and generators conventions, as mentioned above, you'll get correct results.
  2. Apply KCL at the single node and therefore obtain : $$i_c = -i_r$$

Adding an EDIT in order to be more clear and trying to address the user concerns:

If you want a systematic way of solving these exercises, this is probably close to that:

enter image description here enter image description here

In case 1 (in order, LKT, capacitor i,v relationship, and LKC):

$$ \left\{\begin{matrix} v_c = v_r \\ i_c = C\frac{dv_c}{dt} \\ i_c = -i_r = \frac{-v_c}{R} \end{matrix}\right. $$

which gives:

$$\frac{-v}{R} = C\frac{dv_c}{dt}$$

reorder it:

$$\frac{v}{R} + C\frac{dv_c}{dt} = 0$$

Initial and final conditions: $$\left\{\begin{matrix} v_{c0}= V_0 \\ v_{\inf} = 0 \end{matrix}\right.$$

General solution form: $$v_c = A e^{\gamma t} + B$$ Find A and B by applying initial conditions. $$A = V_0, B=0$$ and $$\gamma = -\frac{1}{RC}$$ found from solving the associated equation to the differential equation given:

$$v_c = V_0 e^{-\frac{t}{\tau}}$$

which in turn gives:

$$i_c = - \frac{V_0}{R} e^{-\frac{t}{\tau}}$$

Note that the current will be negative for t>= 0, meaning that it is actually flowing in the opposite direction of the one assumed.

In case 2 instead, the initial conditions are different due to the different conventions adopted. But LKC and LKT still hold as long as you have kept the right conventions for both current and voltage:

$$ \left\{\begin{matrix} v_c = - v_r \\ i_c = C\frac{dv_c}{dt} \\ i_c = i_r = \frac{v_c}{R} \end{matrix}\right. $$

which still gives the exact same differential equation as in case 1 but with different initial conditions since the voltage at time 0 was measured as positive in the "opposite direction":

$$\left\{\begin{matrix} v_{c0}= -V_0 \\ v_{\inf} = 0 \end{matrix}\right.$$

solving as in case 1 gives:

$$v_c = -V_0 e^{-\frac{t}{\tau}}$$

which in turn gives:

$$i_c = \frac{V_0}{R} e^{-\frac{t}{\tau}}$$ now always positive for t>=0.

BIG WARNING NOTE:: This procedure is ok for a simple RC network however when solving a more complex RC network there is a more general procedure to follow which includes this particular easy case. Look for it in a circuit theory book such as "Basic Circuit Theory" by Charles A. Desoer and Ernest S. Kuh.

  • \$\begingroup\$ If I choose in circuit the direction of iC opposite to that on schematic and derive the v(t) I'll get v(t) = e^(t/(R*C))*A but the right result must be with "-" sign in power of "e". How I can understand from here that it's an error? I get the result and if don't look to real answer I'll think that's all are ok. \$\endgroup\$
    – MaxMil
    Dec 19, 2017 at 9:47
  • \$\begingroup\$ Look as the question addition. What do you think. Where I not properly use the textbook method? \$\endgroup\$
    – MaxMil
    Dec 19, 2017 at 12:56
  • \$\begingroup\$ Great info. Please, look at an addition to my question from words "For more clarification..." \$\endgroup\$
    – MaxMil
    Dec 21, 2017 at 7:57

With the current directions shown, the capacitor equation must be set up assuming the capacitor is charging, thus:

\$v=\small V_{C0}+\frac{1}{C}\large \int\small i_C\:dt\$ ... (1)

For the resistor:

\$\small v=R\:i_R \$

and, since \$\small i_R=-i_C\$, we may write:

\$\small v=-R\:i_C \$ ... (2)

Equating (1) and (2)

\$\small V_{C0}+\frac{1}{C}\large \int\small i_C\:dt\:=-R\:i_C\$


\$ \:i_C+\frac{1}{RC}\large \int\small i_C\:dt\:=-\large \frac{V_{C0}}{R}\$


\$\large\frac{di_C}{dt}+\frac{1}{RC}\small i_C=0\$

Let \$\small\tau=RC\$ and solve by integrating factor method:

\$\small IF= e^{t/\tau}\$

\$i_C\: e^{t/\tau} =\large\int\small 0\:dt+A \$

\$i_C=A e^{-t/\tau} \$

Initial condition: at \$\small t=0\$, \$i_C=-\frac{V_{C0}}{R}\$, hence \$\small A=-\large\frac{V_{C0}}{R}\$


$$i_C=-\frac{V_{C0}}{R} e^{-t/\tau} $$

That is, \$i_C\$ flows in the opposite direction to that assumed.

  • \$\begingroup\$ Look the update of the question. I know that your solution is working great but I want to use that from textbook and find the trick that not allow to use it always as I need. \$\endgroup\$
    – MaxMil
    Dec 19, 2017 at 12:56
  • \$\begingroup\$ I don't really understand what you mean; but the reason that the current is shown entering the capacitor in the original question is to reinforce to students, the fact that it doesn't matter a jot which way you choose for the current - the analysis always sorts it out and tells you the actual magnitude and direction \$\endgroup\$
    – Chu
    Dec 19, 2017 at 14:25
  • 1
    \$\begingroup\$ If you assume the current flows into a capacitor, you are assuming it's charging, so \$v=\small V_{C0}+\frac{1}{C} \int idt\$. If you assume current is flowing out of a capacitor, you are assuming it's discharging, so \$v=\small V_{C0}-\frac{1}{C} \int idt\$. Then do the maths and if any of the currents is negative, it means the assumed direction should be reversed; if a current is positive, the assumed direction is correct. \$\endgroup\$
    – Chu
    Dec 19, 2017 at 19:04
  • 1
    \$\begingroup\$ You are very confused about the sign of the exponent. Perhaps you need to read more about solving 1st order diff equations. \$\endgroup\$
    – Chu
    Dec 19, 2017 at 19:10
  • 1
    \$\begingroup\$ Because it's an academic exercise to teach you that it doesn't matter which direction you select as long as your currents all comply with KCL. I've told you this previously. \$\endgroup\$
    – Chu
    Dec 20, 2017 at 9:03

What you thought was absolutely correct,you know that the voltage of capacitor is decreasing so why don't you substitute \$-\mathrm{d}v/\mathrm{d}t\$ in the expression in 7.4b according to the textbook?


I am currently facing this same problem (im still holding my pen 😁), and none of the answers are satisfying to me (maybe im too dumb for those answers) so i tried different approach.

I thought it in terms of physical process, i.e., current is going to decrease via resistor so, with below fig. as reference,

i is going to decrease, so i = - Cdv/dt ( or i = -v/R ) and there we have it, our missing little fella, "The Negative Sign" 😁enter image description here


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