RC circuit

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Given a closed circuit with a source of constant voltage V\,, an ideal capacitor with capacitance C\, and a total equivalent resistanc R\, (which may include the internal resistances of the voltage source, wires and capacitor), we analyse the circuit as follows.

Assign a direction to the unknown time-varying current I(t)\, and use Kirchoff's voltage law. Summing voltage increases around the loop we have:

V - \frac{Q(t)}{C} - I(t)R = 0

Where Q(t) is the time-dependent charge accumulated on the capacitor. Since I(t) = \frac{dQ(t)}{dt}, we can write a single first-order differential equation:

\frac{dQ(t)}{dt}\frac{1}{Q(t)-VC} = -\frac{1}{RC}

which we easily solve by directly integrating from some initial time t_0\, to some final time t_1\,.

ln\left|Q(t_1)-VC\right| - ln\left|Q(t_0)-VC\right| = \frac{t_0 - t_1}{RC}

This is re-arranged into:

Q(t_1) = VC + (Q(t_0)-VC)e^{-(t_1-t_0)/RC}

In the special case of charging a capacitor from zero charge Q(t_0) = 0\, at t_0 = 0\,, we get the familiar result:

Q(t) = VC\left(1-e^{-t/RC}\right)\,.

Similarly for a capacitor with initial charge Q_0\, at t_0 = 0\, and no other voltage sources V = 0\,, we obtain the other familiar result for a discharging capacitor:

Q(t) = Q_0 e^{-t/RC}\,.

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