Question

(a) Find the charge (in C) stored on each capacitor in the figure below (C1 = 12.0 µF, C2 = 4.50 µF) when a 1.53 V battery is connected to the combination.

(b) What energy (in J) is stored in each capacitor?

Answer 1

you didn't give any figure and  third capacitor so i take 0.3 μF

net capacitance of C1 and C2 is C1+C2 as they are in parallel.

so capacitance = C = 12 + 4.50 = 16.50 μF.

so potential difference across C = V1 = 0.3*1.53/(16.50+0.3) = 0.0273 V

[voltage is divided in the inverse ratio of capacitances]

as C1 and C2 are parallel, this is the voltage across both of them

a)

charge across a capacitor is given by q = C*V

so charge on C1

Q1 = C1*V1 = 12*10^-6 * 0.0273 = 0.327 * 10^-6 C

charge on C2

Q2 = C2*V1 = 4.50*10^-6 *0.0273 = 0.122 * 10^-6 C

voltage across 0.3 μF = 1.53 - 0.0273 = 1.50 V

charge across it = 0.3*10^-6 * 1.50 = 0.45*10^-6 C

b)

energy stored in a capacitor is given by (1/2)*C*V ²

so energy in C1

E = (1/2)*12*10^-6 * 0.0273² = 4.478*10^-9 J

energy in C2

E = (1/2)*4.5*10^-6 *0.0273² = 1.67*10^-9

energy in 0.3 μF capacitor = (1/2)*0.3*10^-6 *1.50² = 0.3387*10^-6 J