Worked exercise — capacitor network

Problem

Three capacitors $C_1 = 4\;\mu\text{F}$, $C_2 = 6\;\mu\text{F}$, $C_3 = 3\;\mu\text{F}$ are connected as follows: $C_1$ in series with the parallel combination of $C_2$ and $C_3$. The total voltage is $V = 12\;\text{V}$. Find the equivalent capacitance, the charge on $C_1$ and the voltage across $C_2$ and $C_3$.

$C_1$ in series with the parallel of $C_2$ and $C_3$.

Solution

Step 1 — parallel of $C_2$ and $C_3$ (capacitances in parallel add): $C_{23} = C_2 + C_3 = 6 + 3 = 9\;\mu\text{F}$

Step 2 — series of $C_1$ and $C_{23}$ (in series the reciprocals add): $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_{23}} = \frac{1}{4} + \frac{1}{9} = \frac{9 + 4}{36} = \frac{13}{36} \;\Rightarrow\; C_{\text{eq}} = \frac{36}{13} \approx 2{,}77\;\mu\text{F}$

Step 3 — charge on the series combination. In series the capacitors carry the same charge $Q$: $Q = C_{\text{eq}}\,V = \frac{36}{13}\times 12 = \frac{432}{13} \approx 33{,}2\;\mu\text{C}$ This is the charge on $C_1$ (and on $C_{23}$).

Step 4 — voltages. $V_1 = \frac{Q}{C_1} = \frac{432/13}{4} = \frac{108}{13} \approx 8{,}31\;\text{V}$ $V_{23} = V - V_1 = 12 - \frac{108}{13} = \frac{156 - 108}{13} = \frac{48}{13} \approx 3{,}69\;\text{V}$ Since $C_2$ and $C_3$ are in parallel, they both have the same voltage $V_2 = V_3 = V_{23} \approx 3{,}69\;\text{V}$.

Check: $V_1 + V_{23} = 8{,}31 + 3{,}69 = 12\;\text{V}$. Correct.

$\ev{C_{\text{eq}} \approx 2{,}77\;\mu\text{F},\quad Q_1 \approx 33{,}2\;\mu\text{C},\quad V_2 = V_3 \approx 3{,}69\;\text{V}}$

Links

Topics: Electric circuits Concepts: Capacitance and capacitors Skills: Network simplification