What if the circuit contains components that are neither purely in series nor in parallel?

Use delta-wye (Y-Δ) transforms for capacitors, or apply Kirchhoff's loop and junction rules directly to charges and voltages to form a system of linear equations.

How do you handle capacitors with an initial charge?

Treat the initial charge as an independent voltage source $V_0 = Q_0/C$ in series with an uncharged capacitor, then apply Kirchhoff's rules.

What happens if a dielectric is inserted into one capacitor after the circuit is charged?

If the battery remains connected, the voltage is constant; recalculate the equivalent capacitance and new total charge. If disconnected, total charge is constant; recalculate equivalent capacitance to find the new voltages.

How to solve a capacitor circuit problem

Solving a capacitor circuit requires reducing the network to a single equivalent capacitance, computing the total charge, and expanding backward to find individual charges and voltages. This method applies to direct current (DC) circuits in steady-state conditions where capacitors act as open circuits and hold constant charge.

Capacitors in parallel share voltage and their capacitances add directly. Capacitors in series share charge and their equivalent capacitance is the reciprocal of the sum of reciprocals.

The setup

Identify nodes and branches in the circuit diagram to determine exactly which capacitors are in series and which are in parallel. Series capacitors have no branching between them and thus share the same charge. Parallel capacitors are connected across the same two nodes and thus share the same voltage.

The steps

Identify combinations of parallel capacitors and replace them with an equivalent capacitance: $C_{p} = C_1 + C_2 + \dots + C_n$ .
Identify combinations of series capacitors and replace them with an equivalent capacitance: $\frac{1}{C_{s}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$ .
Repeat steps 1 and 2 until the entire circuit is reduced to a single total equivalent capacitance, $C_{eq}$ .
Calculate the total charge supplied by the source voltage $V$ using $Q_{total} = C_{eq}V$ .
Work backward through your reduced circuits. For series components, set their charge equal to the branch charge and calculate voltage using $V = Q/C$ . For parallel components, set their voltage equal to the node voltage and calculate charge using $Q = CV$ .

Checking the result

Verify Kirchhoff's Voltage Law (KVL): the sum of voltages across series capacitors must equal the total voltage applied across that branch. Verify Conservation of Charge: the sum of charges on parallel capacitors must equal the total charge entering that parallel junction.

Common errors

Using resistor combination rules instead of capacitor rules (adding series directly). Forgetting to invert the reciprocal sum when calculating series capacitance. Assuming series capacitors split voltage equally when their capacitances differ. Assuming parallel capacitors have the same charge.

Worked example

A 12 V battery is connected to a circuit containing $C_1 = 4 \, \mu ext{F}$ in series with a parallel combination of $C_2 = 2 \, \mu ext{F}$ and $C_3 = 6 \, \mu ext{F}$ . Find the charge and voltage on each capacitor.

First, find the equivalent capacitance of the parallel pair $C_2$ and $C_3$ : $C_{23} = C_2 + C_3 = 2 \, \mu ext{F} + 6 \, \mu ext{F} = 8 \, \mu ext{F}$

Next, find the total equivalent capacitance of $C_1$ in series with $C_{23}$ : $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_{23}} = \frac{1}{4 \, \mu ext{F}} + \frac{1}{8 \, \mu ext{F}} = \frac{3}{8 \, \mu ext{F}}$ $C_{eq} = \frac{8}{3} \, \mu ext{F} \approx 2.67 \, \mu ext{F}$

Calculate the total charge supplied by the 12 V battery: $Q_{total} = C_{eq} V = \left(\frac{8}{3} \, \mu ext{F}\right)(12 \, ext{V}) = 32 \, \mu ext{C}$

Since $C_1$ and $C_{23}$ are in series, they both store $Q_{total}$ : $Q_1 = 32 \, \mu ext{C}$ $Q_{23} = 32 \, \mu ext{C}$

Find the voltage across $C_1$ : $V_1 = \frac{Q_1}{C_1} = \frac{32 \, \mu ext{C}}{4 \, \mu ext{F}} = 8 \, ext{V}$

Find the voltage across the parallel combination $C_{23}$ : $V_{23} = \frac{Q_{23}}{C_{23}} = \frac{32 \, \mu ext{C}}{8 \, \mu ext{F}} = 4 \, ext{V}$

Since $C_2$ and $C_3$ are in parallel, they share this voltage: $V_2 = 4 \, ext{V}$ $V_3 = 4 \, ext{V}$

Finally, calculate the individual charges on $C_2$ and $C_3$ : $Q_2 = C_2 V_2 = (2 \, \mu ext{F})(4 \, ext{V}) = 8 \, \mu ext{C}$ $Q_3 = C_3 V_3 = (6 \, \mu ext{F})(4 \, ext{V}) = 24 \, \mu ext{C}$

Check: $Q_2 + Q_3 = 8 \, \mu ext{C} + 24 \, \mu ext{C} = 32 \, \mu ext{C} = Q_{23}$ . $V_1 + V_{23} = 8 \, ext{V} + 4 \, ext{V} = 12 \, ext{V}$ .

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References: University Physics with Modern Physics (14th Edition), Chapter 24 · Fundamentals of Physics (Halliday & Resnick, 10th Edition), Chapter 25