Question

Four capacitors are connected in series across a 12.0-V battery. Their capacitances are 1.00, 2.00, 4.00 and 4.00 mF. What must be the capacitance of a fifth capacitor that when added in series to the circuit reduces the voltage across the 1.00 mF capacitor by 1.00 V?

Answer 1

V = Voltage of the battery = 12 volts

Total capacitance is given as

1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄

1/C_total = 1/1 + 1/2 + 1/4 + 1/4

C_total = = 0.5 mF

Q₁ = charge through C₁ = C_total V = (0.5 x 10^-3) 12 = 6 x 10^-3 C

V₁ = Voltage across C₁ = Q₁/C₁ = 6 x 10^-3 /(1 x 10^-3) = 6 volts

V₁' = reduced Voltage = 5 volts

Q₁' = new charge in the circuit = C₁ V₁' = (1 x 10^-3) 5 = 5 x 10^-3 C

V = 12 volts

new total capacitance , C_total' = Q₁'/V = (5 x 10^-3 )/12 = 0.42 x 10^-3

new total capacitance is given as

1/C_total' = 1/1 + 1/2 + 1/4 + 1/4 + 1/C

1/(0.42 x 10^-3) = 1/1 + 1/2 + 1/4 + 1/4 + 1/C

C = 0.00042