Find the following. (Let C1 = 15.00 µF and C2 = 9.00 µF.) (a) the equivalent cap

ID: 1535310 • Letter: F

Question

Find the following. (Let C1 = 15.00 µF and C2 = 9.00 µF.) (a) the equivalent capacitance of the capacitors in the figure above µF

(b) the charge on each capacitor on the right 15.00 µF capacitor µC

on the left 15.00 µF capacitor µC

on the 9.00 µF capacitor µC

on the 6.00 µF capacitor µC

on the left 15.00 µF capacitor V

on the 9.00 µF capacitor V

on the 6.00 µF capacitor V

6.00 AuF 9.00 V

6 uF and C2 are parallel.

so, C' = 6 + 9 = 15 uF

this is in series with C1 and C1

so, 1 / Ceq = 1 / 15 + 1/ 15 + 1/ 15

Ceq = 5 uF

(b)

The capacitors C1, C' and C1 are in series so, the charge on these capacitors is same .

Q = Ceq*V = 5*9 = 45 uC

charge on parallel capacitors,

When the capacitors are connected in parallel then the potential difference is constant while the charge varies.

Voltage across both capacitors V' = Q / C

V' = 45 / 15 = 3 V

charge on 6 uF ,

Q1 = 6*3 = 18 uC

charge on C2

Q2 = 9*3 = 27 uC

(c)

potential difference across right and left 15 uF,

V1 = V2 = 45 / 15 = 3 V

potential difference across 9 uF,

V3 = 45 / 15 = 3 V

6 uF and 9 uF capacitors are in parallel so, the potential difference across the capacitors is same.

V3 = V4 = 3 V

answer

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