Applying KVL on RC circuit to get the natural reponse v_c(t>0)

Question

I want to find the equation for the discharging of the capacitor \$C\$. My initial thought was to use KVL in the closed loop that is the circuit with the current \$i\$ going anti-clockwise (since the capacitor has \$+\$ on top the current will go anti-clockwise). This gives the equation

\$-v_C + v_R = 0 \Leftrightarrow \$

\$-v_C + Ri = 0 \Leftrightarrow \$

\$RC \frac{dv_c}{dt} = v_C \Leftrightarrow \$

\$\frac{dv_C}{dt} = \frac{1}{RC}v_c\$

Solving this gives: \$v_c(t) = v_c(0) e^{\frac{t}{RC}}\$

However, this is exponential growth and will go to infinity as t goes to infinity. What I want to get is \$v_c(t) = v_c(0) e^{-\frac{t}{RC}}\$ which tends to \$0\$.

My school book use the upper branch as a node and applies KCL to it with currents going downwards for both the resistor and the capacitor. This will give the correct answer.

What I want to know is what have I done wrong in my method with KVL that gives the wrong answer?

Answer 1

Your mistake was substituting: $$i= +CdV_c/dt$$ This would have been true if the current was flowing "into" the capacitor (charging). But here the capacitor is discharging, and hence current direction is in the opposite direction. Therefore: $$i = - CdV_c/dt$$

Now you will reach the given solution.

Answer 2

The current through the capacitor is \$+C\frac{dV_C}{dt}\$ clockwise. Therefore if you go anti-clockwise, like what you did, the current in the circuit would be \$-C\frac{dV_C}{dt}\$. Replace this in your third equation and you will come up with the right equation.

Answer 3

http://www.electronics-tutorials.ws/rc/rc_1.html

Set Vs=0V and you will get the correct answer