Electrical – charging capacitors in series discharging in parallel

Consider this circuit:

^{simulate this circuit – Schematic created using CircuitLab}

I know you didn't specify a resistor in the circuit. Its purpose will become clear later.

Let's say that initially \$V_{C1} = 1V\$ and \$V_{C2} = 0V\$. The charge in C1 is:

$$ Q_{C1} = CV = 1F \cdot 1V = 1C $$

The total energy in the circuit is the same as the energy in C1, because there is no other stored energy elsewhere in the circuit:

$$ E_{C1} = \frac{Q^2}{2C} = \frac{(1C)^2}{2F} = 0.5J $$

When the switch is closed, some current flows. The total charge in the circuit must remain the same, and we can see that the voltage across the capacitors must be equal once the circuit reaches equilibrium.

$$ Q_{C1} = Q_{C2} = 0.5C $$

$$ V_{C1} = V_{C2} = \frac{Q}{C} = \frac{0.5C}{1F} = 0.5V $$

The energy in the capacitors is:

$$ E_{C1} = E_{C2} = \frac{(0.5C)^2}{2F} = 0.125J $$

We have two of these capacitors so the total energy is twice that, 0.25J. Initially we had 0.5J. Where did we lose half the energy?

Consider that in the instant the switch was closed, there is 1V across R. The current is thus 1V/R. The power is thus:

$$ P = EI = 1V \cdot \frac{1V}{R} = \frac{(1V)^2}{R} $$

As you decrease R, the power goes up, approaching infinity:

$$ \lim_{R \searrow 0} \frac{(1V)^2}{R} = \infty $$

Thus, the lost energy was lost as heat in R. The energy lost is the same for any value of R. R can't be made equal to 0Ω without resulting in infinite power, which is impossible.

Incidentally, this is why charge pumps can't be 100% efficient.

Let's translate your words into an equation for the equivalent impedance.

Since C1 and R1 are in parallel

$$Z_1 = R_1||\frac{1}{sC_1}$$

likewise C2 and R2

$$Z_2 = R_2||\frac{1}{sC_2}$$

Solving three impedances in series

$$Z_E = Z_1 + Z_2 + R_3$$