Capacitors in series

Question

A 2μF and a 4μF capacitor are connected in series. A voltage of 1 kV is applied to the series capacitors. After charging, the voltage source is disconnected, and the capacitors are connected to each other so that the terminals of like signs are joined together. Find the charge on each capacitor and the voltage across each capacitor.

Answer 1

Let

C₁ = 2μF = 2×10^-6 F be the capacitance of the first capacitor,

C₂ = 4μF = 4×10^-6 F be the capacitance of the second capacitor,

V = 1kV = 10³V be the voltage applied.

Capacitors connected in series store the same amount charge and act like

one capacitor with capacitance C satisfying

C = C₁C₂/(C₁ + C₂)

In the initial circuit, the amount of charge Q stored in the effective capacitance C

(as well as on each capacitor) is

Q = VC

After the circuit is reconnected as described,

the charges get redistributed with the result that the capacitors have some charges Q₁ and Q₂ respectively.

By Kirchoff's law the capacitors must have the same voltage.

So we have

Q₁/C₁ = Q₂/C₂

And by conservation of charge

Q₁ + Q₂ = Q = VC₁C₂/(C₁ + C₂)

Substituting Q₂ from the first equation into the second:

Q₁ + Q₁C₂/C₁ = VC₁C₂/(C₁ + C₂)

Hence

Q₁ = VC₁²C₂/(C₁ + C₂)²

Substituting actual numbers

Q₁ = 10³×(2×10^-6)²×(4×10^-6)/(2×10^-6 + 4×10^-6)² = 4/9 mC

Since C₂ = 2C₁, we have Q₂ = 2Q₁ = 8/9 mC

The common voltage across the two capacitors is

Q₁/C₁ = (4/9)×10^-3/2×10^-6 = 2/9 kV