Question

A 0.41 μF and a 0.65 μF capacitor are connected in series to a 15 V battery.

Part A] Calculate the potential difference across each capacitor.

Part B] Calculate the charge on each capacitor.

Part C] Repeat part A assuming the two capacitors are in parallel.

Part D] Repeat part B assuming the two capacitors are in parallel.

Answer 1

Remember:
For parallel combination
Ceq = C1 + C2 + C3 +...............
for series combination
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ............
for 2 capacitors in series it will be
Ceq = C1*C2/(C1+C2)
Using this Information:

Part A.

When both capacitors are connected in series, then

Ceq = C1*C2/(C1 + C2)

Ceq = 0.41*0.65/(0.41 + 0.65)

Ceq = 0.25 F

Qeq = Ceq*V

Qeq = 0.25*10^-6*15

Qeq = 3.75 C

Now remember in capacitors parallel combination voltage distribution in each part will be same and in series combination charge distribution in each capacitor will be same.

Since both are in series, So

Q1 = Q2 = 3.75*10^-6 C

V1 = Q1/C1 = 3.75*10^-6/(0.41*10^-6) = 9.15 V

V1 = Voltage across 0.41 F capacitor = 9.2 V

V2 = Q2/C2 = 3.75*10^-6/(0.65*10^-6) = 5.78 V

V2 = Voltage across 0.65 F capacitor = 5.8 V

Part B

Q1 = charge across 0.41 F capacitor = 3.75*10^-6 C = 3.75 C

Q2 = charge across 0.65 F capacitor = 3.75*10^-6 C = 3.75 C

Part C.

when both capacitors are in parallel, then Voltage distribution in each part will be same and equal to total voltage

V1 = Voltage across 0.41 F capacitor = 15 V

V2 = Voltage across 0.65 F capacitor = 15 V

Part D.

Q1 = C1*V1 = 0.41*10^-6*15 = 6.15*10^-6 C

Q1 = charge across 0.41 F capacitor = 6.15 C

Q2 = C2*V2 = 0.65*10^-6*15 = 9.75*10^-6 C

Q2 = charge across 0.65 F capacitor = 9.75 C

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