a) If you have five capacitors with capacitances 1.1 × 10 -6 F, 1.7 × 10 -6 F, 4

Question

a) If you have five capacitors with capacitances 1.1 × 10-6 F, 1.7 × 10-6 F, 4.5 × 10-6 F, and two 9.7 × 10-6 F in series. What is the equivalent capacitance of all five?

b) Initially the capacitors are uncharged. Now a 12 V battery is attached to the system. How much charge is on the positive plate of the 4.5 × 10-6 F capacitor?

c)What is the potential difference between the plates of the 4.5 × 10-6 F capacitor?

d) How much energy is stored in the entire capacitor system?

e) If you have five capacitors with capacitances 1.1 × 10-6 F, 1.7 × 10-6 F , 4.5 × 10-6 F, and two 9.7 × 10-6 F in parallel. What is the equivalent capacitance of all five?

f) If one attaches a 12 V battery to the system, how much charge is on the positive plate of the 4.5 × 10-6 F capacitor?

g)What is the potential difference between the plates of the 4.5 × 10-6 F capacitor?

h) How much energy is stored in the entire capacitor system?

Explanation / Answer

A) for capacitors in series

1/Cnet = (1/C1)+(1/C2)+(1/C3)+(1/C4)+(1/C5)

1/Cnet = (1/1.1)+(1/1.7)+(1/4.5)+(1/9.7)+(1/9.7)

Cnet = 0.519*10^-6 F

B) In series combination,the same amount of charge is present on all plates of capacitor

Qnet = Cnet*V = 0.519*10^-6*12 = 6.23*10^-6 C

C) Potential difference across 4.5 *10^-6 F is V = Qnet/C = (6.23*10^-6)/(4.5*10^-6) = 1.38 V

D) U = 0.5*Cnet*V^2 = 0.5*0.519*10^-6*12^2 = 3.74*10^-5 J

E) in parallel

Cnet = C1+C2+C3+C4+C5

Cnet = (1.1+1.7+4.5+9.7+9.7)*10^-6 = 26.7 *10^-6 F

F) same potential difference across any capacitor = 12 V

then charge on 4.5*10^-6 F is Q = C*V = 4.5*10^-6*12 = 54*10^-6 C

G) potential difference is 12 V

H) U = 0.5*C*V^2 = 0.5*26.7*10^-6*12^2 = 192.24*10^-5 J

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