C 1 = 1.00 ? F and C 2 = 9.00 ? F.) (a) the equivalent capacitance of the system
ID: 1337534 • Letter: C
Question
C1 = 1.00 ?F
and
C2 = 9.00 ?F.)
(a) the equivalent capacitance of the system
(b) the charge on each capacitor
(c) the potential difference across each capacitor
For the system of capacitors shown in the figure below, find the following. (Let CI-1.00 F and C2 9.00 F.) Ci 6.00 F 2.00 F C2 90.0 V (a) the equivalent capacitance of the system (b) the charge on each capacitor on C on C2 on the 6.00 F capacitor on the 2.00 F capacitor HC JIC AC AC (c) the potential difference across each capacitor across C1 across C2 across the 6.00 F capacitor across the 2.00 F capacitor Need Help? Read ItExplanation / Answer
given ,
C1 = 1.00 F
C2 = 9.00 F
a)
In a series capacitor,
1/C = 1/1 + 1/6
C = 0.857 µF
1/C' = 1/9 + 1/0.857
C' = 0.783 µF
parallel capacitor,
C total = C + C'
C total = 0.857 + 0.783
C total = 1.64 µF
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b)
Q1 = Capacitence * Voltage
Q1 = (0.857 * 90)
Q1 = 77.13 µC
charge on C1 is 77.13 µC
6 µF capacitor is 77.13 µC
Q2 = C' * V
Q2 = (0.783 * 90)
Q2 = 70.47 µC
charge on C2 is 70.47 µC
2 µF capacitor is 70.47 µC
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c)
potential difference at C1 is
V1 = Q1/C1
V1 = (77.13) / (1)
V1 = 77.13 V
potential difference at 6 µF is
V1' = Q1/C
V1' = (77.13) / (6 )
V1' = 12.855 V
potential difference at C2 is
V2 = Q2/C2
V2 = (70.47) / (0.857 )
V2 = 7.83 V
potential difference at 2 µF is
V2' = Q2/C
V2' = (70.47) / (2 )
V2' = 35.235 V
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