C 1 = 1.00 ? F and C 2 = 9.00 ? F.) (a) the equivalent capacitance of the system

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 It

Explanation / 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

===========================================

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

==================================================

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

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.