a) If you have five capacitors with capacitances 0.8 × 10 -6 F, 2.3 × 10 -6 F, 3

Question

a) If you have five capacitors with capacitances 0.8 × 10-6 F, 2.3 × 10-6 F, 3.7 × 10-6 F, and two 7.6 × 10-6 F in series. What is the equivalent capacitance of all five?

C =

b) Initially the capacitors are uncharged. Now a 14 V battery is attached to the system. How much charge is on the positive plate of the 3.7 × 10-6 F capacitor?

Q =

c)What is the potential difference between the plates of the 3.7 × 10-6 F capacitor?

V=

How much energy is stored in the entire capacitor system?

PE =

can anyone help?

Explanation / Answer

a)

1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4 + 1/C5

1/Ceq = 1/(0.8*10^(-6)) + 1/(2.3*10^(-6)) + 1/(3.7*10^(-6)) + 1/(7.6*10^(-6)) + 1/(7.6*10^(-6))

1/Ceq = 1.25*10^6 + 0.4347*10^6 + 0.27*10^6 + 0.1316*10^6+ 0.1316*10^6 = 2.218*10^6

Ceq = 0.45*10^(-6) F = 0.45 uF

b)

)In series the charge is the same on each capacitor...so using Q = Ceq*V

Q = Ceq*V

Q = 0.45*10^(-6)*14 = 6.3*10^(-6) C

c)

V = Q/C = 6.3*10^(-6)/(3.7*10^(-6)) = 1.7 V

d)

PE = 1/2*Ceq*V^2 = 0.5*0.45*10^(-6)*(14)^2 = 4.41*10^(-5) J

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