I understand that something that is charged with one coulomb and has a potential difference of one volt across it will have a capacitance of one farad. \$C = \frac{q}{V}\$. I get that because current is the derivative of charge with respect to time, \$I = C\frac{dV}{dt}\$.
This is where I start to get confused. If you divide both sides by \$C\$ and integrate with respect to time, shouldn't that just give you \$V = \frac{q}{C}\$?
What I don't get is how this becomes \$V(t) = \frac{1}{C}\int_{t_{0}}^{t}I(\tau)\, d\tau + V(t_{0})\$.
This is saying that \$q = \int_{t_{0}}^{t}I(\tau)\, d\tau + V(t_{0})\$. I don't get why, though.
Shouldn't \$q = \int I\, dt\$ and that be the end of it? I mean, obviously this isn't the case since my textbook and everything I can find online says otherwise, but I don't get it.
Best Answer
q is not equal to \$ \int_{t0}^{t} I(\tau) d\tau + V(t_0) \$
it has no sense to sum apples (charge) and pears (potential) :)
if you have \$ V(t)=\frac{1}{C}\int_{t0}^{t} I(\tau) d\tau + V(t_0) \$
\$ \frac{1}{C} \$ is just multiplying \$\int_{t0}^{t} I(\tau) d\tau \$ and not \$ V(t_0)\$
\$ \int_{t0}^{t} I(\tau) d\tau \$ is a \$ \Delta Q\$
so \$ V(t)=\frac{1}{C}\int_{t0}^{t} I(\tau) d\tau + V(t_0)=\frac{\Delta Q}{C} +V(t_0) \$
I hope this is what you were missing :)