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# Capacitor on breadboard

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.
And now we're ready too.

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.