1 answer

A.) Find the charge stored on each capacitor in the figure shown above(C1=11.1 uF, C2= 9.62uF)...

Question:

a.) Find the charge stored on each capacitor in the figure shown above(C1=11.1 uF, C2= 9.62uF) When a 1.57 V battery is connected to the combination.
Q1=
Q2=
Q3=

b.) What energy is stored in each capacitor?
E1=
E2=
E3=

0.300 uF ne charge stored on each capacitor in the figure shown above (C1 = 11.1 pF, C2 = 9.62 pF) when a 1.57 V

0.300 uF ne charge stored on each capacitor in the figure shown above (C1 = 11.1 pF, C2 = 9.62 pF) when a 1.57 V

Answers

from the cirucit c1 and c2 are in parallel combination

so c12 = c1+c2 = 11.1*10^-6+9.62*10^-6 = 20.72*10^-6 F

c12 and 0.3*10^-6 F are in series then

c tot = 0.3*10^-6*20.72*10^-6/(0.3+20.72)*10^-6

c tot = 0.296*10^-6 F

total charge q tot = c tot*v = 0.296*10^-6*1.57 = 0.465*10^-6 C

in a series combination charge is same across the capacitors

q12 = q3 = 0.465*10^-6 C

voltage across c12 is v12 = 0.465*10^-6/20.72*10^-6

v12 = 0.0224 V

voltage across c3 is v3 = 0.465/0.3 = 1.55 V

charges on c1 and c2

q1 = 11.1*10^-6*0.0224 = 0.249*10^-6 C

q2 = 9.62*10^-6*0.0224 = 0.215*10^-6 C

b) energy

E1 = q1^2/2C1 = (0.249*10^-6)^2/2*11.1*10^-6 = 2.79*10^-9 J

E2 = q2^2/2c2 = (0.215*10^-6)^2/2*9.62*10^-6 = 2.4*10^-9 J

E3 = (0.465*10^-6)^2/2*0.3*10^-6 = 0.36*10^-6 J

.

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