What is the voltage across each capacitor in this circuit?

[thisisfudd]

If 26 V is applied across the following network, calculate the voltage across each capacitor:

Ignore the dots, it's just spaces between the two ends

___ 3.00 uF _____ 4.00 uF
|.......|
|.......|
__________2.00 uF______
|.......|
|.......|
|.......|

So I know that for capacitors in series, it's the opposite of resistors, ie it's the reciprocal, and in parallel it's the opposite, ie you just add them. So I have voltage across a whole network. I assume I just need to do Q = CV? But how do I add these capacitances together. Maybe that's the question I'm really asking. Ie, how to I add one parallel to two in series?

 

[StatusX]

The voltage across the top two will equal the voltage across the bottom one. You can use the rule for adding capacitances to get the total capacitance of the top two, and from that get the total charge stored. This is a little confusing to interpret, though.

Here's what it means. The important point is that there is no way for extra charge to get in between the capacitors, so whatever charge is on the inner plate of one, the opposite charge is on the inner plate of the other. Then, use the fact that the charge on the plates of a single capacitor are equal in magnitude and opposite in sign to get the charge on the outer plates. Now the charge you calculated from the total capacitance as I mentioned above is just the charge stored on these outer plates. You can easily see that this is also equal to the charge stored on each plate, as is the case for any number of capacitors in series.

 

[thisisfudd]

So that means that the total capacitance is 7?

 

[StatusX]

No, the rule is 1/C_T = 1/C₁ + 1/C₂. The total capacitance of the circuit is then C_T + C₃, but this won't help you get the charge on each capacitor. Like I said, use the fact that the total voltage across C₁ and C₂ is the same as the voltage across C₃.

 

[leolaw]

After we found the total capacitance of the whole circuit, we can get the charge that come out from the battery by using Q = VC.
Since that charges are conserved, is the sum of the amount of charge from the top 2 capacitances and the bottom capitance equal to the charges come out from the battery?