# Applying KCL to RC circuit and then solving DE for voltage

When you derive the voltage equation for an RC circuit using KCL, they normally show both resistor current $i_r$ and capacitor current $i_c$ as leaving the same node. In reality the current is getting sourced from the discharging capacitor and flows in one direction.

However, when I use the later case I get exponential growth instead of exponential decay when i solve the DE for voltage. Why is that?

Case 1:

$$i_c + i_r = 0$$

$$C\frac{dV}{dt} + \frac{V}{R} = 0$$

$$v~dv = -\frac{1}{RC}~dt$$

$$\int \limits_{v(0)}^{v(t)} \frac{1}{v} ~dv = -\frac{1}{RC}~\int \limits_{0}^{t}d\tau$$

$$ln(v_1/v_0) = \frac{-1}{RC}(t - 0)$$

$$v(t) = v(0)e^{-t/(RC)}$$

exponential delay. good.

Case 2:

$$i_c - i_r = 0$$

$$C\frac{dV}{dt} - \frac{V}{R} = 0$$

$$v~dv = \frac{1}{RC}~dt$$

$$\int \limits_{v(0)}^{v(t)} \frac{1}{v}~dv = \frac{1}{RC}~\int \limits_{0}^{t}d\tau$$

$$ln(v_1/v_0) = \frac{1}{RC}(t - 0)$$

$$v(t) = v(0)e^{t/(RC)}$$

I remember in school many years ago they said you can assign the current arrows anyway you want, its just the sign comes out minus if you assign the direction wrong, and it corrects itself.

I'm wondering how to make case 2 have exponential delay, the same as case 1.

• First sort out the equations, there are lots of errors, e.g. $\int\: v\:dv =\frac{v^2}{2}$ – Chu Jan 10 '20 at 0:05
• @TonyStewartSunnyskyguyEE75, Yes, but the maths is arbitrary and the OP needs to formulate the problem properly. – Chu Jan 10 '20 at 0:09
• yes of course. I agree now The convention for voltage polarity must follow chosen direction of convention for current and if reversed so must negate polarity for voltage. – Tony Stewart Sunnyskyguy EE75 Jan 10 '20 at 0:14
• see this answer: electronics.stackexchange.com/questions/296866/… – Big6 Jan 10 '20 at 16:17
• take a look at this one too (your same question): electronics.stackexchange.com/questions/253433/… – Big6 Jan 10 '20 at 16:18