Capacitors in parallel: what is the total capacitance when multiple capacitors are connected in parallel (assume ideal components)?

Introduction / Context:Capacitors are often combined to achieve target values or to improve ripple performance. Knowing how capacitance combines in series vs. parallel is vital for power supplies, filters, and timing circuits.

Given Data / Assumptions:

• Two or more ideal capacitors connected in parallel.
• Same node voltage across each branch (by parallel definition).
• No frequency-dependent parasitics considered for the rule (ESR/ESL ignored).

Concept / Approach:Charge stored by a capacitor is Q = C * V. In parallel, each capacitor sees the same voltage and stores Q_i = C_i * V. Total charge is Q_total = Σ Q_i = V * Σ C_i, so the equivalent capacitance satisfies C_eq = Q_total / V = Σ C_i. Therefore, capacitances add directly in parallel.

Step-by-Step Solution:Write charge relation: Q_i = C_i * V (same V across all branches).Sum charges: Q_total = Σ C_i * V.Divide by V: C_eq = Σ C_i.Therefore, the total capacitance equals the sum of individual capacitances.

Verification / Alternative check:Numerical example: 10 μF ∥ 22 μF gives C_eq = 32 μF, matching the addition rule and common design practice in decoupling networks.

Why Other Options Are Wrong:Sum of capacitive reactances: reactance is frequency-dependent; also, the rule is about capacitance, not reactance.Source voltage / total current: defines impedance at a given frequency, not a direct rule for capacitance.Less than smallest capacitance: that applies to series capacitance, not parallel.

Common Pitfalls:Confusing series and parallel rules; mixing frequency-dependent reactance with capacitance; ignoring ESR/ESL in high-frequency designs.

Final Answer:equal to the sum of the individual capacitance values