gmmstr827 said:
(a)
Right... so... find the individual charges without changing anything first?
Q1 = (2.70 µF)(475 V) = 1282.5 µC
Q2 = (4.00 µF)(525 V) = 2100 µC
Now add them together.
Q = Q1+Q2 = 1282.5 µC + 2100 µC = 3382.5 µC
(b)
So I "turn around" the plate by switching C1 and C2, right? You never commented on this.
C1 = 4.00 µF
C2 = 2.70 µF
Now using the total charge and each capacitor, I can find the voltages.
V=Q/C
V1 = 3382.5 µC / 4.00 µF
V1 = 845.63 V
V2 = 3382.5 µC / 2.70 µF
V2 = 1252.78 V
The question now asks to find the charges... again.
Q=VC
Q1= V1*C1 = 3382.52 µC
Q2= V2*C2 = 3382.51 µC
They have the same charges, different only through rounding error.
If this isn't right, please just show me how you would do it.
For (a), you're mostly on target. You have the total charge which is the sum of the charges which were on the individual capacitors before they were connected together. Now, after they are connected in parallel, they are sharing that total charge. They are now two capacitors in parallel with a total charge of Q = 3382.5 µC.
The capacitance of the parallel pair is Ceq = C1+C2 = 6.7 µF. The voltage across Ceq is given by V1 = Q/Ceq, or 504.9 V. So that's the voltage that the paralleled capacitors have on them.
Both of the individual capacitors of that parallel pair have that voltage across them -- they
must have the same voltage because they are wired in parallel.
To determine the individual charges that the capacitors now have on them (remember that the total charge from before the connection of the capacitors in parallel is now spread over the two parallel-connected capacitors), you use the V = Q/C formula for each of the capacitors, where V is now the new shared voltage of the two capacitors, V1, that we just determined.
For part (b), There is no "turning around" a plate. The problem states that the capacitors are connected so that the negative and positive plates of the opposing capacitors are wired together, rather than positive and positive, negative negative as in part (a). Shuffling capacitor variable names is not going to accomplish this. Here is what is happening:
Take a 2.70 µF capacitor in your right hand. It has two leads. Let's paint one lead red and the other lead black, strictly for purposes of identification. Now, the red lead is connected to the positive terminal of a 475V supply. The black lead is connected to the negative terminal of the same supply. After a short time, you remove the capacitor's connections from the supply. The capacitor is now charged to 475V and holds 1283.5 µC of charge, which it retains after disconnection from the supply. Now you take a 4.00 µF capacitor in your left hand. You give it the same treatment as the other capacitor, only this time you connect it to a 525V supply. It ends up with 2100 µC stored and 525V across its (painted) leads.
The plot so far:
C1: 2.70 µF ; 475V ; 1283.5 µC ; red lead connects to plate with +++ charges ; black lead, --- charges
C2: 4.00 µF ; 525V ; 2100.0 µC ; red lead connects to plate with +++ charges ; black lead, --- charges
Now you connect the capacitors in parallel. The "switch" that's described in the problem statement involves how the leads of the capacitors are connected. In this case, the positive (red) lead from each capacitor is connected to the negative (black) lead of the other. The capacitors are in parallel, but they have been wired together in such a way as to connect their oppositely charged plates via their leads. Red to black, black to red.
What happens? The opposite charges residing on the plates of the individual capacitors suddenly "see" a path to opposite charges via the newly connected leads. They jump at the chance to meet, and promptly race together, joining and canceling each other out. This happens on both plates of each capacitor simultaneously; the charges on both plates of each capacitor have access to the charges on the plates of the other capacitor, and they are charges of the opposite sign.
When all is said and done (in a few nanoseconds or so), after the mad rush of charges between capacitors, all the charges that can cancel have done so. What remains is whatever charges did not find an opposite charge to cancel with. So whichever capacitor initially had the largest charge "wins"; It's remaining charges determine the whole charge for the parallel combination. Those remaining charges spread themselves across the parallel capacitors just as in part (a). With the remaining charge you can determine the voltage across the equivalent capacitance of the paralleled capacitors. With the voltage, you can determine how much of the remaining charge resides on the individual capacitors of the pair.
So in this case C2 contained the greater charge (2100 µC) before the capacitors were connected. The remaining charge after the mutual annihilation will be Q2 - Q1. Because C2 had the greater charge, it will determine the voltage polarity of the paralleled result; C2's red lead will be connected to the positively charged plates of the paralleled pair.
So to sum up for part (b). Capacitors C1 and C2 with charges 1283.5 and 2100 µC and 475 and 525 volt potential differences are connected with opposing plate polarity. The charges on their plates meet, greet, and annihilate, leaving Q2 - Q1 = 817.5 µC to populate the plates of the now parallel pair. Their parallel capacitance of 6.7 µF then sports a voltage of (Q2 - Q1)/(C1 + C2) = 122V. Both of the capacitors, because they are in parallel, now share this voltage. The charges that reside on each capacitor can be found via the Q = C*V formula. The sum of the charges will equal the "remains", Q2 - Q1.
How's that?
