(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?
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
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