Bonus track for survivors. Mathematical skills!

Did you ask yourself why charging/discharging graphs have that kind of time behavior?
Integral operation has the answer (symbol: $\int$). What does it do?
Even if we consider them almost stationary - no change in time - in reality we're treating time-dependent sizes; assuming that this sizes change every infinitesimal instant, we use integral operation to rebuild the whole behavior.

Help yourself thinking you're on a roll-coaster, watching constantly to ground, trying to measure the distance at every moment from starting to ending position; now sum all your computations and integral result is ready.

So considering our initial circuit operating with constant sizes (not varying in time), and assuming that starting C voltage is zero we can write equations in which power supply is equal to sum of voltage-differences both on R and C: $V=\Delta V_{R} + \Delta V_{C} = \Delta V_{R} + \frac{\Delta Q}{C}$

We derive (other mathematical operation) first and last equation members; in more human words we examine the equation on single generic quick instants, same kind of above "infinitesimal instant": $0 = \delta V_{R} + \delta V_{C} = R\frac{di(t)}{dt} + \frac{i(t)}{C}$

Constant size like our power supply V are zero after derivation, because this measures the size's variation degree: a constant simply doesn't vary.

Now we have a first-order differential equation with constant coefficients.
Resolving it means that we rebuild the whole current behavior: $\int_{0}^{t}\frac{1}{i(\tau)}di(\tau) = -\frac{1}{RC}\int_{0}^{t}d\tau \Longrightarrow ln|i(t)|_{0}^{t} = -\frac{1}{RC}\tau|_{0}^{t} \Longrightarrow ln\frac{i(t)}{i(0)} = -\frac{1}{RC}t \Longrightarrow i(t) = i(0)e^{-\frac{t}{RC}}$

We found mathematical explanation of current behavior. What about voltage?
Consider that a voltage present on an electrical item at a certain instant is given by:

1. starting voltage $V_{C0}$ (zero for our capacitor);
2. remaining voltage charge for C.

Translated becomes $v_{C}(t) = \Delta V_{C} + V_{C0}=\frac{1}{C}\int_{0}^{t}e^{-\frac{\tau}{RC}}d\tau + 0$

We know explicit expression of i(t) so substituting we get $v_{C}(t) = \frac{i(0)}{C}\int_{0}^{t}e^{-\frac{\tau}{RC}}d\tau = -i(0)\cdot Re^{-\frac{\tau}{RC}}|_{0}^{t} = V_{0}(1-e^{\frac{t}{RC}})$

Finally we can justify exponential behavior of both voltage and current in capacitor charging (similar with discharging)

• $i(t) = i(0)e^{-\frac{t}{RC}}$
• $v(t) = V_{0}(1-e^{\frac{t}{RC}})$

[Tip: inserting C before R you get same results, but only for constant sizes (in DC, direct coupling); in AC (alternate coupling) this is not more possible.]

Here we're playing with quite low component sizes (capacitor used in test is 10 micro-Farad, $10^{-5}\mu F$; not so negligible indeed) and you can get charged C in your hands safely: generally it's recommended to take care and get capacitors with protection or not by its terminals at least.