1. The problem statement, all variables and given/known data A potential difference of V = 38.0 V is applied across a circuit with capacitances C1 = 4F, C2 = 5F, and C3= 2F, as shown in the figure. What is the magnitude and sign of q3l, the charge on the left plate of C3 (marked by point A)? What is the electric potential difference, V3, across C3? What is the magnitude and sign of the charge q2r, on the right plate of C2 (marked by point B)? 2. Relevant equations Q = CV Useful concepts: Charge for capacitors in series is the same. Current for resistors/capacitors in series is the same. Voltage for resistors/capacitors in parallel is the same. 3. Solution: 1. C1+C2=C12 = 9F Cequivalent = (C12 * C3)/C12 + C3) = ~1.6F q3l = Ceq * ∆V = 60.8V 2. V3 = Q3/C3 = 60.8V = 30.4 V 3. q2r = C2/(C1 +C2 ) * Q3 = 34.05 C Already have the solutions, just need explanations. C1 and C2 are in parallel, therefore should have the same Voltage (potential difference). Third one is the one I'm having trouble with. It's so similar to the "resistance of interest equation", like in here, when finding V_1. It's so frustrating, I know all the formulas, Q = CV, capacitance in series just add up (e.g. C12 + C3). But no one ever tells how to find different voltages / charges on different elements on the circuit. Like, in this example, would you still be able to use the "resistance of interest" formula: R1/(R1+R2) * Vs (supplied voltage) if additionally further down the circuit there were a bunch of other resistors connected in series or neither in series nor in parallel? Would you have to account for them when finding Voltage (V1) across R1? ----- 1. The voltage across C1 and C2 is the same, correct? But their sum doesn't equal 38V, nor does it mean that voltage across C1 is 38V, correct? But whatever the voltage across C1 is, it's of the same magnitude in C2, despite the fact that capacitors have different capacitance? 2. Why is the equivalent capacitance and the given voltage is used to calculate the charge on C3 instead of Q3 = C3 * V3? I know the formula Q3 = C3 * V3 is valid, but how come the equivalent capacitance * given voltage also yields V3? Because at the left plate of C3, the current have traveled far enough to have passed all capacitors (hence the equivalent capacitance), and at that point voltage must therefore be the same as it was supplied initially? So you treat the circuit at that point as if it has one capacitor (Ceq)? Or because in the end, the same voltage must return to the negative end of Voltage supply...? And because of the KVL loop rule, that voltages must add to 0...? But I thought Voltage either rises or drops across a capacitor. Once current passes either through a capacitor or a resistor, once it leaves an element, it will have different Voltage... uhm? 3. How does one figure the Voltage across C2? Like, how do you know how much of the supplied voltage would go to both V1 and V2? Is it even correct to say: "go to"? I mean, Voltage has a relationship with current. And the more current goes to a capacitor, the higher the voltage will be, right?