Problem — network of capacitors

Problem

Network of capacitors. Three capacitors $C_1 = 3\ \mu\text{F}$, $C_2 = 6\ \mu\text{F}$, $C_3 = 4\ \mu\text{F}$ are connected as in the figure ($C_2 \parallel C_3$ in series with $C_1$):

Diagram: $C_1$ in series with the parallel combination of $C_2$ and $C_3$, powered by the source $\mathcal{E}$.

(a) Calculate the equivalent capacitance across the source. (b) With $\mathcal{E} = 12\ \text{V}$ determine the charge on $C_1$. (c) Compare the total energy $\tfrac12 C_\text{eq}\mathcal{E}^2$ with the sum $\tfrac12 C_i V_i^2$ of each capacitor.

Solution

(a) Equivalent capacitance. In parallel the capacitances add: $C_{23} = C_2 + C_3 = 6 + 4 = 10\ \mu\text{F}$ In series the inverses add: $\frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_{23}} = \frac{1}{3} + \frac{1}{10} = \frac{13}{30}\ \mu\text{F}^{-1} \;\Rightarrow\; C_\text{eq} = \frac{30}{13} \approx 2{,}3\ \mu\text{F}$

(b) Charge on $C_1$. In a series combination all elements carry the same charge, equal to the charge delivered by the source: $Q_1 = C_\text{eq}\,\mathcal{E} = 2{,}31 \cdot 10^{-6} \cdot 12 \approx 2{,}8 \cdot 10^{-5}\ \text{C}$

(c) Energy balance. The voltages across the two series blocks: $V_1 = Q_1/C_1 = 2{,}77\cdot10^{-5}/3\cdot10^{-6} \approx 9{,}23\ \text{V}$, and across the parallel combination $V_{23} = \mathcal{E} - V_1 \approx 2{,}77\ \text{V}$. Total energy from the source: $U_\text{tot} = \tfrac12 C_\text{eq}\mathcal{E}^2 = \tfrac12 \cdot 2{,}31\cdot10^{-6} \cdot 12^2 \approx 1{,}66\cdot10^{-4}\ \text{J}$ Sum over the individual capacitors ($C_2$ and $C_3$ have the same $V_{23}$): $\sum \tfrac12 C_i V_i^2 = \tfrac12(3\cdot10^{-6})(9{,}23)^2 + \tfrac12(6\cdot10^{-6}+4\cdot10^{-6})(2{,}77)^2 \approx 1{,}28\cdot10^{-4} + 0{,}38\cdot10^{-4} = 1{,}66\cdot10^{-4}\ \text{J}$ The two values coincide: the energy stored in the network is the sum of the energies of the individual capacitors, as required by energy conservation.

$\ev{C_\text{eq} \approx 2{,}3\ \mu\text{F} \qquad Q_1 \approx 2{,}8\cdot10^{-5}\ \text{C} \qquad U_\text{tot} \approx 1{,}66\cdot10^{-4}\ \text{J}}$

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Topics: Circuiti elettrici Concepts: Capacità e condensatore Skills: Semplificazione di reti