Posted by Becky on Tuesday, April 1, 2008 at 1:37pm.
If I understand the circuit, the 3.0 and 4.0 uF capacitors are connected in series with a 10.0 uF capacitor connected in parallel and all of this across a 40 v cell. First determine the capacitance of the series 3.0 and 4.0.
1/C = 1/3.0 + 1/4.0 = 1.7 uF.
Then add in the parallel 10.0 to make a total of 11.7 uF.
Q = Coulombs of stored electricity = CV where C is capacitance (in F) and V is voltage. Check my thinking.
For the 10 uF capacitor,
Q' = C' V = (10*10^-6)*40 = 400 uC
For the series 4 uF and 3 uF capacitors, which both acquire the same charge Q",
10 V = Q"/4*10^-6 + Q"/3*10^-6
= Q"/1.714*10^-6
The two series capacitors act like a single 1.714 uF capacitor
Q" = 17.14 uC is the charge on both the 3 uF and 4 uF capacitors
I forgot to use 40 V for the series capacitors. Their charge is 4 times higher than what I stated. Also, the battery only has to deliver a charge of Q", not the sum of the charges on the two series capacitors, which would be 2Q"
DrBob222 already provided a correct answer by computing an effective C for all three capacitors in series-parallel
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