Four capacitors are connected as shown in the figure below. (C = 12.0 µF.) http:

Question

Four capacitors are connected as shown in the figure below. (C = 12.0 µF.)
http://www.webassign.net/serpse8/26-p-023-alt.gif
(a) Find the equivalent capacitance between points a and b.


µF

(b) Calculate the charge on each capacitor, taking ?Vab = 20.0 V.
20.0 µF capacitor

µC
6.00 µF capacitor


Find the voltage drop across the 20.0-µF capacitor and use that result to find the voltage drop across the 6.00-µF capacitor. µC
3.00 µF capacitor


You appear to have calculated the charge on the 3.00-µF capacitor in terms of the charges you calculated in earlier parts of the problem, but since at least one of those earlier values was incorrect, the charge you calculated for the 3.00-µF capacitor is also incorrect. µC
capacitor C µC

Explanation / Answer

1/(1/12E-6+1/3E-6)2.4 F+6 F=8.4 F==>1/(1/8.4E-6+1/20E-6)

a) 5.92 F=Ceq

b) q=C*V==>20 V*5.92 F=118.31 C total==>V20F=118.31 C/20 F=5.92 V

(20-5.92)=14.08 V for the parallel circuit==>14.08 V*2.4 F=33.8 C==>

V6F=14.08 V==>33.8 C/12 F=2.82 V==>33.8 C/3 F=11.26V=V3F

12 F=2.82 V drop==>3 F V drop=11.26 V==>6 F V drop 14.08 V

20 F V drop=5.92 V  

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