How can I solve this? Thanks in advance!

Question

In (Figure 1), let V = 15.0 V and C1=C2=C3= 24.2 μF.

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

Answer 1

Dr Gulshan solved section A for this problem.

Here are solutions for sections B and C

B) For capacitors in series the equivalent Capacitance Ceq is found from this 1/Ceq = 1/C1 + 1/C2 +1/C3

1/ Ceq = C2C3 + C1C3+ C1C2 / C1C2C3 but since all capacitors are the same then C = C1 =C2 =C3 =24.2μF

1/Ceq = C²+C²+C²/ C³ = 3C²/C³ = 3/C, and Ceq = C/3 = 24.2μF/3 = 8.07 μF

Find Voltage for each capacitor first find Q = Ceq V = 8.07 μF ( 15 V) = 121.05 x10^-6 C

V1 = Q / C1 , V2 = Q / C2, V3 = Q / C3 but since C1 =C2 =C3 then the voltage in each capacitor is

V = Q/C = 121.05x10^-6 C / 24.2 x 10^-6 F = 5.0 Volts, to check this the sum of all voltages in the capacitors should equal the source voltage V = 15 volts. For three identical capacitors each carrying 5 volts the sum of voltages is 3(5volts)= 15 volts

The total energy stored in the system is E_T = E1 + E2 + E3, where E = 1/2 CV² since C and V are the same for all capacitors E_T = 3 E = 3 ( 1/2 )(24.2μF)(5V)² = (3/2)(24.2x10^-6F)(5V)² = 907.5 x10 ^-6Joules

= 907.5 x 10^-3 mJ = 0.908 mJ

C) For capacitors in parallel, the voltage is the same such that V = V1 = V2 = V3

The charge in each capacitor is Q = CV since C and V are the same for all capacitors

The charge stored per capacitor is Q = 24.2μF (15V) = 24.2 x10^-6F ( 15V) = 363 x 10^-6 C

The total charge stored = 3( 363 x 10^-6 C) = 1089 x 10^-6 C

This could also be obtained by realizing that Ceq for capacitors in parallel

Ceq= C1+C2+C3 = 3( 24.2μF) = 72.6 μF

and Q = Ceq (V) = 72.6 μF ( 15 V) = 1089 x 10^-6 C

The Total Energy stored in all capacitors is therefore

E_T = 1/2 ( Ceq)(V)² = 1/2 (72.6μF)(15 V)² = (1/2)(72.6 x 10^-6F)(15 V)² = 8167.5 x 10^-6 Joules

= 8167.5 x 10^-3 mJ = 8.17 mJ

Answer 2

Hello Alyssa,

In This question C2 and C3 are in series and the two are then in parallel with C1

The Equivalent capacitance is

C1 +( C2*C3 ) / (C1 +C2) = 24.2 +585.64/48.4 = 24.2 +12.1 = 36.3 μF = 3.63* 10 ^-5 F

Enery stored in whole network = 1/2 ( CV² ) = 0.5 * 3.63*10 ^-5 *15 *15 = 408.38 * 10 ^-5 J = 4.08*10 ^-3 J