Capacitors in Series

Series capacitors carry the same branch current, and the same charge change passes through each. Their equivalent capacitance is smaller than either capacitor. C1 is 15 µF and C2 is 30 µF.

$$\begin{aligned} \frac{1}{C_{\mathrm{eq}}} &= \frac{1}{C_1}+\frac{1}{C_2} \\ C_{\mathrm{eq}} &= \frac{C_1C_2}{C_1+C_2} =10\,\mu\mathrm{F} \\ \tau &= R_1C_{\mathrm{eq}} =10\,\mathrm{ms} \end{aligned}$$

Read the circuit

V1 drives the capacitor pair through 1 kΩ. The source rises from 0 to 5 V at 5 ms, holds high for 50 ms, then returns to zero. Both capacitors start uncharged.

• Vc — green: total voltage across the pair.
• Vc2 — red: voltage across the lower capacitor C2, measured relative to ground.
• I — blue: common charging current through R1, C1, and C2.

The voltage across C1 is $V_{C1}=V_c-V_{C2}$. The midpoint is not ground, so its ground-referenced probe reads C2's voltage, not C1's.

Why the smaller capacitor gets more voltage

For the same charge, voltage is inversely proportional to capacitance. With zero initial charge on the isolated midpoint:

$$\begin{aligned} V_{C1} &= \frac{C_2}{C_1+C_2}V_c =\frac{2}{3}V_c \\ V_{C2} &= \frac{C_1}{C_1+C_2}V_c =\frac{1}{3}V_c \end{aligned}$$

After a long 5 V pulse, C1 approaches 3.33 V and C2 1.67 V. They each hold a charge magnitude of $Q=C_{\mathrm{eq}}V_c=50\,\mu\mathrm{C}$, although their voltages differ.

Observe in CircuitSim

At about 15 ms, one time constant after the rising edge, Vc is about 3.16 V, Vc2 1.05 V, and I 1.84 mA. Just before 55 ms, Vc is about 4.97 V. After the falling edge, current reverses while the capacitor voltages remain continuous and decay toward zero.

The saved transient uses Use initial conditions, with both initial capacitor voltages set to zero. The isolated midpoint has no DC path, so its starting state is specified explicitly. The 0–110 ms run and 50 µs maximum step show both charging and discharging. Open Table for Time, Vc, Vc2, and I.

Try two changes

1. Set C1 → Capacitance to 30u (30 µF). Equal capacitors split the voltage equally. Their equivalent is 15 µF and the time constant is 15 ms; each approaches 2.5 V.
2. Restore C1 to 15u, then set R1 → Resistance to 2k. The time constant becomes 20 ms and the initial charging current is approximately 2.5 mA. The final voltage ratio is unchanged. Restore R1 to 1k afterward.

Model assumptions

The capacitors are ideal, with no leakage, ESR, or voltage limits. Different initial charges or leakage paths can change series voltage sharing. Real series capacitor stacks may require balancing resistors; simply adding voltage ratings does not guarantee safe sharing.