For a normal RC circuit with one resistor and one capacitor, let say at $t=0$ the switch closes and complete the circuit. All components are connected. voltage source (VA), $R$ and $C$.
If C was initially charged to let say a voltage value of -VC(initially) before the switch even closed to complete the circuit and now when $t=0$ the source is trying to charge the capacitor as VA > -VC(initially)
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Wouldn't the capacitor have to discharge to zero voltage first before it can be charged up to the source voltage?
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How does the discharge of the capacitor and the charging work at the same time? I can't visualize the concept of how that work. Where does the energy go?
To my understand,
q=c*v; I=dq/dt
E – IR – V(c) =0
E- (dq/dt)R – q/c =0 Solving the DE equation: Vc(t) = Vf (1-e^-1/RC); where vf is when the capacitor has been charged for long time and it is now act as an open circuit. No current flow, which mean the resistor voltage is zero and the sum of the voltage drop across the circuit is now just E = Vf (final voltage across the capacitor equal to the source).
I(t) = Io*e^(-t/RC);
the Io (I initial) is when t=0 the switch just closed. The charge across the capacitor is still zero as it can't change instantaneously. therefore, the initial current Io is just the Io=E/R. However this is not what we have in this case.
Best Answer
Yes.
It doesn't work at the same time; the capacitor is either supplying energy to the circuit (discharging) or receiving energy from the circuit (charging).
As I've written here and at the EE stackexchange site, the general solution for the switched RC circuit is (for $t \ge 0$)
$$v_C(t) = V_S + \left[v_C(0) - V_S\right]e^{-t/RC}$$
where $V_S$ is the source voltage and $v_C(0)$ is the initial capacitor voltage. The capacitor current is then
$$i_C(t) = C\frac{dv_C}{dt} = \frac{1}{R}\left[V_S - v_C(0)\right]e^{-t/RC} $$
The instantaneous power delivered to the capacitor is then
$$p_C(t) = v_C(t) \cdot i_C(t) = \frac{V_S}{R}\left[V_S - v_C(0)\right]e^{-t/RC} - \frac{\left[v_C(0) - V_S\right]^2}{R}e^{-2t/RC}$$
Now, let $v_C(0)$ (the initial capacitor voltage) be negative. Note carefully that the power delivered to the capacitor is initially negative, crosses zero, reaches a maximum positive value, and then decays to zero.
When the power is negative, the capacitor is discharging, supplying energy to the circuit (the resistor receives the energy initially stored in the capacitor). When the power is positive, the capacitor is charging, receiving energy from the circuit.
To get a feel for how this capacitor power equation works, follow the link to a Desmos Graphing Calculator page that I created
For further work, use the capacitor power formula above to see what happens if the capacitor voltage is initially greater than the source voltage.
Easier to just derive it here than to search for it.
KVL:
$$v_R = V_S - v_C$$
Ohm's Law:
$$i = \frac{V_S - v_C}{R}$$
Capacitor equation:
$$i = C\frac{dv_C}{dt}$$
$$\Rightarrow \frac{dv_C}{dt} + \frac{1}{RC}v_C = \frac{V_S}{RC}$$
Homogeneous solution:
$$v_C(t) = Ae^{-t/RC}$$
Particular solution:
$$v_C(t) = V_S$$
$$\Rightarrow v_C(t) = V_S + Ae^{-t/RC}$$
$$v_C(0) = V_S + A \rightarrow A = v_C(0) - V_S$$
$$\therefore v_C(t) = V_S + \left[v_C(0) - V_S \right]e^{-t/RC}$$