Question

A 0.50-?F and a 1.4-?F capacitor (C1 and C2, respectively) are connected in series to a 15-V battery.

Part A. Calculate the potential difference across each capacitor. Express your answers using two significant figures separated by a comma.

Part B. Calculate the charge on each capasitor. Express your answers using two significant figures separated by a comma.

Part C. Calculate the potential difference across each capacitor assuming the two capacitors are in parallel. Express your answers using two significant figures separated by a comma.

Part D. Calculate the charge on each capasitor assuming the two capacitors are in parallel. Express your answers using two significant figures separated by a comma.

Answer 1

(a)
Charge on a capacitor is Q = CV
If you have two capacitors(C1 and C2) the charge across them is Q₁ = C₁*V₁ and Q₂ = C₂*V₂
Two capacitors in series must hold the same charge so Q₁ = Q₂
C₁*V₁ = C₂*V₂
V₁ = C₂*V₂/C₁
By Kirchhoff's voltage law: V₁ + V₂ = 15,
plug in V₁ from the previous step:
C₂*V₂/C₁ + V₂ = 15
1.4*V₂/0.5 + V₂ = 15
V₂ = 3.95 V
V₁ = C₂*V₂/C₁ = 1.4*3.95/0.5 = 11.05 V

(b)
Q₁ = C₁*V₁, Q₂ = C₂*V₂
Q₁ = 0.5 x 10^-6*11.05 = 5.525 x 10^-6 C
Q₂ = 1.4 x 10^-6*3.95 = 5.525 x 10^-6 C

c) If you have components in parallel, the voltage across them is the same. In this case if you have a 15V battery in parallel with two capacitances, then the voltage across each capacitor MUST be 15V also.

For capacitors in parallel, the charges are NOT the same(although they may be if the capacitances are equal, which they're not for this question).
Since V1 = V2 = 15, then
Q1 = C1*V1 = C1*15 and
Q2 = C2*V2 = C2*15