Four capacitors are connected as shown in the figure below. (Let C = 18.0 µF.)
Find the equivalent capacitance between points a and b
Calculate the charge on each capacitor, taking ΔVab = 17.0 V
Q = CΔV
C = Q / ΔV
ΔV1 = ΔV2 = ΔVtot (for parallel capacitors)
Q tot = Q1 + Q2 (charge on parallel capacitors, it says)
ΔVtot = ΔV1 + ΔV2 (for series capacitors)
Q = Q1 = Q2 (charge on series capacitors, it says)
The Attempt at a Solution
The only thing I could solve was the equivalent capacitance across, which is 6 x 10-6
And I broke it up like this (and it's correct):
I have NO IDEA at all how to determine the individual charge on each of them, even with the formulas...
I can see ΔV across series is supposed to add up to 17V but no idea how.
Where should I start on that?
Edit: ok used Q = CV for the (20 x 10-6 f) capacitor and got (102 x 10-6 C) because charge is supposed to be the same across series
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