Applying KCL to RC circuit and then solving DE for voltage

Question

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?

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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.

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Case 2:

$$i_c - i_r = 0$$

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

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

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

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

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

??Exponential growth?? bad.

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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.

Answer 1

In case 2, you reversed the direction of the capacitor current. Thus, the current is no longer given as CdV/dt but is now -CdV/dt. When you make that correction, the equation becomes the same as for case 1 and hence has the same solution, as it should. Remember, you can assign the current directions any way you want but then you must follow the circuit laws when calculating voltages.