# Discharge of a capacitor

by Ahmedbasil
Tags: capacitor, discharge
 P: 8 1. The problem statement, all variables and given/known data You have a capacitor, of capacitance C farads, with charge Q coulombs. It is connected in series with a resistor of resistance R ohms. Derive an expression for the potential difference over the capacitor at any time t. 2. Relevant equations and theorems $$I_{c} = C\frac{dV_{c}}{dt}$$ $$V = IR$$ *Kirchhoff's voltage law 3. The attempt at a solution Using KVL: $$V_{c} - I_{c} R = 0$$ $$V_{c} = RC\frac{dV_{c}}{dt}$$ $$\frac{1}{RC} dt = \frac{1}{V_{c}} dV_{c}$$ then: $$V_{c} (t) = V_{initial} e^{\frac{t}{RC}}$$ where $$V_{initial} = \frac{Q}{C}$$ 3. My concern as t approaches infinity the potential difference over the capacitor also approaches infinity. This is definitely not right - the capacitor is discharging. Every textbook/website I look at comes up with the equation: $$V_{c} (t) = V_{initial} e^{\frac{-t}{RC}}$$ For the life of me I cannot figure out what I'm doing wrong. I know what's supposedly wrong with my solution, but I cannot see the mathematical proof of the minus sign on the power of e. I'd greatly appreciate any help or insight anyone could give.
 P: 8 I suppose that dVc/dt doesn't direction into account. Why doesn't it though? At first I considered this, but then I dismissed it, knowing that it's perfectly fine to have a rate of change w.r.t another variable negative. The equation should be: $$I_{c} = C |\frac{dV_{c}}{dt}|$$ Anyway, thanks very much for your help.
 Quote by Ahmedbasil I suppose that dVc/dt doesn't direction into account. Why doesn't it though? At first I considered this, but then I dismissed it, knowing that it's perfectly fine to have a rate of change w.r.t another variable negative. The equation should be: $$|I| = C |\frac{dV_{c}}{dt}|$$ Anyway, thanks very much for your help.