Capacitor Question for Physics 2

Three capacitors C1 = 1.5 µF, C2 = 9.5 µF, and C3 = 14.5 µF are connected as shown with a battery of voltage V = 18 V. (When entering units, use micro for the metric system prefix µ.)

http://www.webassign.net/ncsuplseem2/3-post-002.gif

(a) What is the equivalent capacitance of this circuit?
I got this answer to be 1.4 microF which was correct.


(b) What is the charge on C1?


(c) What is the voltage across each capacitor?
V1 =
V2 =
V3 =

(d) What is the charge on C2 and C3?
Q2 =
Q3 =

a) C2 parallel with C3 ---> Cp = 9.5 + 14.5 = 24 µF
The equivalent capasitors :
Ce = (C1 * Cp)/(C1 + Cp)
= ( 1.5 * 24)/(1.5 + 24) = 1.41 µF

b) the charge on C1 = Qe
Qe = Ce * V
= 1.41 * 18 = 25.38 µC

c) V1 = Qe/C1 = 25.38/1.5 = 16.92 V
V2 = Qe/Cp = 25.38/24 = 1.06 V
V3 = V2 = 1.06 V

d) Q2 = C2 * V2 = 9.5 * 1.06 = 10.07 µC µC
Q3 = C3 * V3 =14.5 * 1.06 = 15.37

(b)

Charge of global equivalent capacitor (at 1.412µF), is
Qceq=V x Ceq but across Ceq you get the voltage battery:
so Qceq=18 x 1.4E-06= 2.54E-05 C


(c)
In serie capacitors: the current is the same for each capacitor, charge is the same:
Qceq= Qc1= Qc2//3

(c2//3 being the equivalent capacitor of C2 and C3)


Thanks to this, you can calculate voltage across each capacitor:

Vc1= Qceq/C1= 16.9V
According to mesh law, Vc2//3= Vc2 =Vc3, because C2 and C3 are in parallel.

Again accordin to this very law:
Vbatt= Vc1 + Vc2//3
===>18= 16.9 + Vc2//3
===> Vc2//3= Vc2 = Vc3 = 1.1V


(d)
Qc2= Vc2 x C2= 1.1 x 9.5E-06 = 1.045 E-05 C
Qc3= Vc3 x C3= 1.1 x 14.5E-06= 1.595E-05 C