Question

Find the following. (In the figure use C1 = 19.60 μF and C2 =13.60 μF.)



(values are μF)
a) The equivalent capacitance of the capacitors in the figureabove
_____μF
b) The charge on each capacitor
on the right 19.60 μF capacitor_____μC
on the   13.60 µF capacitor ________μC
on the 6.00 µF capacitor__________μC
(c) the potential difference across each capacitor
on the right 19.60 µFcapacitor______V
on the left 19.60 µFcapacitor_______V
on the 13.60 µFcapacitor__________V
on the 6.00 µF capacitor___________V

Answer 1

a)The equivalent capacitance of the capacitors 6.00 μFand C₂ = 13.60 μF is

C' = 6.00 + C₂ = 6.00 + 13.60 = 19.60 μF

The capacitors C₁,C' and C₁ are inseries therefore the equivalent capacitance is

(1/C'') = (1/C₁) + (1/C') + (1/C₁)

C₁ = 19.60 μF

or (1/C'') = (1/19.60) + (1/19.60) + (1/19.60)

or (1/C'') = (3/19.60)

or C'' = (19.60/3) = 6.53 μF

Therefore,the equivalent capacitance of the capacitors in thefigure above is C'' = 6.53 μF

b)The capacitors C₁,C' and C₁ are inseries therefore the charge on these capacitors is same and isgiven by

Q = C'' * V

V = 9.00 V

or Q = 6.53 * 10^-6 * 9.00 = 58.77 * 10^-6C = 58.77 μC

Therefore,the charge on the right 19.60 μF capacitor is58.77 μC

The charge on the capacitor C' is 58.77 μC

The capacitor C' is a parallel combination of capacitors 6.00μF and C₂ = 13.60 μF.When the capacitors areconnected in parallel then the potential difference is constantwhile the charge varies.The total potential difference on the twocapacitors is

Q = C' * V'

or V' = (Q/C') = (58.77 * 10^-6/19.60 *10^-6) = 3.0 V

The charge on the two capacitors is

Q₁ = 6.00 * V' = 6.00 * 10^-6 * 3.0 =18.0 * 10^-6 C = 18.0 μC

and Q₂ = C₂ * V' = 13.60 *10^-6 * 3.0 = 40.8 * 10^-6 C = 40.8μC

(c)The potential difference across the right 19.60 µF capacitor is

V₁ = (Q/19.60 * 10^-6) = (58.77 *10^-6/19.60 * 10^-6) = 3.0 V

The potential difference across the left 19.60 µF capacitor is

V₂ = (Q/19.60 * 10^-6) = (58.77 *10^-6/19.60 * 10^-6) = 3.0 V

The potential difference across the 13.60 µF capacitor is

V₃ = (Q/C') = (58.77 * 10^-6/19.60 *10^-6) = 3.0 V

The 13.60 μF and the 6.00 μF capacitors are inparallel therefore the potential difference across the capacitorsis same that is 3.0 V.