Capacitance Rules and Series Combination Calculations

Equivalent Capacitance Rules for Parallel and Series Combinations

The resultant or equivalent capacitance of any number of capacitors connected in parallel is the sum of their individual capacitances. When capacitors are connected in parallel, each capacitor experiences the same potential difference across its plates, and the total stored charge is equal to the sum of the charges stored across each individual capacitor.

The reciprocal of the equivalent capacitance of any number of capacitors connected in series is the sum of the reciprocals of the individual capacitances. When capacitors are connected in series, the magnitude of charge on each capacitor plate is identical, while the total potential difference applied across the entire combination is equal to the sum of the potential drops across each capacitor.

Example 3.10: Series Capacitor Combination

Two capacitors of capacitances 5 μF5\,\mu\text{F} and 3 μF3\,\mu\text{F} are connected in series and the resulting combination is connected across a potential difference of 600 V600\,\text{V}. The objective is to determine (i) the equivalent capacitance of the combination and (ii) the total charge on the combination.

Series combination of 5 microfarad and 3 microfarad capacitors

To find (i) the equivalent capacitance CC of the combination, apply the reciprocal formula for series capacitors:

1C=15+13=815\frac{1}{C} = \frac{1}{5} + \frac{1}{3} = \frac{8}{15}

Solving for CC yields:

C=158 μF=1.875 μF=1.875×10−6 FC = \frac{15}{8}\,\mu\text{F} = 1.875\,\mu\text{F} = 1.875 \times 10^{-6}\,\text{F}

To find (ii) the total charge qq on the combination, use the formula relating charge, equivalent capacitance, and voltage:

q=CVq = CV

Substituting the values of equivalent capacitance C=1.875×10−6 FC = 1.875 \times 10^{-6}\,\text{F} and voltage V=600 VV = 600\,\text{V} gives:

q=(1.875×10−6×600) Coulombsq = (1.875 \times 10^{-6} \times 600)\,\text{Coulombs}

q=1.125×10−3 Cq = 1.125 \times 10^{-3}\,\text{C}