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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.

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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 :)

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This is really just a math question.

When you take the integral \$ \frac{1}{C}\int_{t_0}^t I(\tau)d\tau \$, you get \$ \frac{Q(t)}{C} - \frac{Q(t_0)}{C} \$ = \$ V(t) - V(t_0)\$. This is bad, because we just want \$ V(t) \$; the simple way of fixing this is to add \$V(t_0)\$ to cancel out the term in the integral.

We definitely can't just write \$ \frac{1}{C} \int I(t) \$ - what are you going to do with the constant of integration?

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