Problem 12: Two capacitors of capacitances 4µF and 8 µF are first connected (a) in series and then (b) in parallel. In each case external voltage source is 200 V. Calculate in each case the total capacitance, the potential drop across each capacitor and charge on each capacitor.

Solution

(a)

We know that when capacitors are connected in series, then

Therefore, the equivalent capacitance in series combination will be

Now the charge Q on each capacitor is, Q = C_eV, therefore,
Q = 2.66 × 10^-6×200 = 532 × 10^-6 = 5.32 × 10^-4C.
To calculate the potential drop across each capacitor, we use V = Q/C

(b)

When the capacitor are connected in parallel, then the equivalent capacitance C_e is
C_e = C₁ + C₂ = 8 + 4 = 12 µF

Voltage across both capacitors connected in parallel is same; the voltage applied.
V₁ = V₂ = 200 V

Charges on the capacitors can be calculated with Q = CV
Q₁ = C₁V = 4 × 10^-6× 200 = 800 × 10^-6 = 800 µC
Q₂ = C₂V = 8 × 10^-6× 200 = 1600 × 10^-6 = 1600 µC