Question

Three capacitors of 2.00 nF, 5.00 nF and 7.00 nF are connected in series to a source

with a potential difference of 9.00 V.

a)What is the equivalent capacitance?

b) What is the the charge in each capacitor, and what is the potential difference across

each capacitor?

c) What is the energy stored in each capacitor?

9) Repeat 8 with the same three capacitors in parallel. How is this different than the

result in series?

Answer 1

C1= 2 nF C2= 5 nF C3= 7 nF Voltage V= 9 Volts

Connected in Series

a)  1/Ceq = 1/C1 + 1/C2 + 1/C3 = 1/2 + 1/5 + 1/7 = 59/70

Ceq = 70/59 =1.186 nF -----Equivalent capacitance in series

b) 1/Ceq = V/Q

Q = V*Ceq = 9 * 1.186 = 10.674 nC ---- Charge is same across all capacitors in series

V= V1+ V2 + V3

V1 = Q/C1 = 10.674/2 = 5.337 Volts

V2= Q/C2= 10.674/5 = 2.134 Volts

V3= Q/C3 = 10.674/7 = 1.524 Volts

c)

energy stored in capacitor 1 = 1/2 C1 V1² = 1/2 * 2*10^-9 * 5.337²  = 28.48 nJ

energy stored in capacitor 2 = 1/2 C2 V2² = 1/2 * 5*10^-9 *2.134 ²  = 11.38 nJ

energy stored in capacitor 3 = 1/2 C3 V3² = 1/2 * 7*10^-9 * 1.524²  = 8.12 nJ

Connected in Parallel:

a) Ceq = C1 + C2 + C3 = 2+5+7 = 14 nF --Equivalent capacitance in parallel

b) Q= Q1+Q2+Q3

and Ceq = Q/V

Voltage remains constant across all capacitors in parallel

Q= Ceq (V) = (C1+C2+C3)*V =

so Q1 = C1V = 2*9 = 18 nC

Q2=C2V= 5*9 = 45 nC

Q3 = C3V= 7*9= 63 nC

c) V1=V2=V3=V= 9Volts

energy stored in capacitor 1 = 1/2 C1 V1² = 1/2 * 2*10^-9 * 9²  = 81 nJ

energy stored in capacitor 2 = 1/2 C2 V2² = 1/2 * 5*10^-9 *9 ²  = 202.5 nJ

energy stored in capacitor 3 = 1/2 C3 V3² = 1/2 * 7*10^-9 * 9²  = 283.5 nJ