Equivalent capacitance of three capacitors

Problem

Synthesis: equivalent capacitance. Three capacitors: $C_1 = 4\;\mu\text{F}$ and $C_2 = 6\;\mu\text{F}$ in parallel with each other, in series with $C_3 = 5\;\mu\text{F}$. (a) Equivalent capacitance. (b) With an applied voltage $V = 12\;\text{V}$, total charge and stored energy.

Solution

(a) Equivalent capacitance. $C_1$ and $C_2$ are in parallel, so the capacitances add: $C_{12} = C_1 + C_2 = 4 + 6 = 10\;\mu\text{F}.$ This block is in series with $C_3$, so the reciprocals add: $\frac{1}{C_\text{eq}} = \frac{1}{C_{12}} + \frac{1}{C_3} = \frac{1}{10} + \frac{1}{5} = \frac{3}{10}\;\mu\text{F}^{-1},$ $\ev{C_\text{eq} = \frac{10}{3}\;\mu\text{F} \approx 3{,}33\;\mu\text{F}}.$

(b) Charge and energy. The total charge delivered by the source flows through the series combination: $Q = C_\text{eq}\,V = 3{,}33\cdot 10^{-6}\cdot 12 \approx \ev{Q = 40\;\mu\text{C}}.$ The stored energy: $U = \frac{1}{2}C_\text{eq}V^2 = \frac{1}{2}\cdot 3{,}33\cdot 10^{-6}\cdot 12^2 \approx \ev{U = 240\;\mu\text{J}}.$

Connections

Topics: Electric field and potential Concepts: Capacitance and capacitor · Field energy density Methods: Series-parallel simplification Objects: Parallel-plate capacitor