Capacitors in Parallel

Parallel capacitors share the same voltage, and their capacitances add. C1 is 4 µF and C2 is 6 µF. Together they store the same total charge as a 10 µF capacitor at the same voltage.

$$\begin{aligned} C_{\mathrm{eq}} &= C_1+C_2 =10\,\mu\mathrm{F} \\ \tau &= R_1C_{\mathrm{eq}} \\ &= (1\,\mathrm{k}\Omega)(10\,\mu\mathrm{F}) =10\,\mathrm{ms} \end{aligned}$$

Read the circuit

V1 sends a 0–5 V pulse through R1. Both capacitor branches connect to the same upper node and lower return. R1 controls the charging current and provides a discharge path when V1 returns to zero.

• Vc — green: the common voltage across C1 and C2.
• Ic1 — blue: current into C1.
• Ic2 — purple: current into C2.

The current through R1 is the sum $I=I_{C1}+I_{C2}$. Positive branch current charges a capacitor; negative branch current means discharge opposite to the arrow.

Why the larger capacitor takes more current

Both capacitors see the same rate of voltage change. Their currents therefore divide in proportion to capacitance:

$$\begin{aligned} I_{C1} &= C_1\frac{dV_c}{dt}=0.4I \\ I_{C2} &= C_2\frac{dV_c}{dt}=0.6I \\ \frac{I_{C2}}{I_{C1}} &= \frac{C_2}{C_1}=1.5 \end{aligned}$$

Just after the rising edge, the total current is nearly 5 mA: approximately 2 mA into C1 and 3 mA into C2. At 5 V, the charge magnitudes approach 20 µC and 30 µC respectively.

Observe in CircuitSim

Transient simulation opens over 0–110 ms with a 50 µs maximum step. The pulse rises at 5 ms and returns to zero at about 55 ms. Vc rises continuously; both branch currents decay as charging slows. At about 15 ms, Vc is 3.16 V, Ic1 0.736 mA, and Ic2 1.10 mA.

After the falling edge, both currents reverse but keep the 2:3 ratio, while Vc decays smoothly. The voltage and current chart groups use separate scales. Open Table for Time, Vc, Ic1, and Ic2. The normal DC operating point starts both capacitors at zero voltage.

Try two changes

1. Set C2 → Capacitance to 4u (4 µF). Both branches take equal current. Total capacitance becomes 8 µF and the time constant becomes 8 ms; each initial charging current is nearly 2.5 mA.
2. Restore C2 to 6u, then set R1 → Resistance to 2k. The time constant doubles to 20 ms; initial currents halve to about 1 mA and 1.5 mA. Restore R1 to 1k afterward.

Model assumptions

These capacitors have no ESR, leakage, or voltage limits. Real parallel capacitors can share fast transient current differently because their ESR and wiring inductance differ. Parallel connection increases capacitance; it does not add their voltage ratings.