They give the following circuit.
Each capacitor is 50*10^-6 [F], the Resistors are 2[Ω], and the Voltage is 5[V].
In the first step the close only the 1st switch and let the capacitor to be charged. I've found that the charge is 2.5*10^-6 [C]. (using the equation Q = V*C)
On the second part the OPEN the 1st switch and CLOSE the 2nd switch, so that the 3rd remains OPEN.
The question what is the charge on each of the capacitors on equilibrium?
Q = V*C
The Attempt at a Solution
First of all, I don't understand what is happening with the Voltage source. Is he in the circuit? because the 1st switch is open
Well, my attempt was that both of the Capacitors are now connected in series so that:
so the total capacitance is (C/2)
so the total charge will be Q = V*(C/2), but it seems weird to me that the total charge will be less than the first capacitor in the beginning.