Series capacitors: what is the correct statement about the equivalent capacitance of a series combination of two or more capacitors?

Introduction / Context:
Designers often place capacitors in series to increase voltage rating or to achieve a specific, smaller equivalent capacitance. Understanding how series combinations behave prevents mistakes in timing circuits, filters, and power electronics where precise capacitance and voltage distribution matter.

Given Data / Assumptions:

• Two or more ideal capacitors are connected in series.
• Leakage and tolerance are ignored for the conceptual rule.
• We focus on the magnitude of equivalent capacitance C_eq.

Concept / Approach:
For capacitors in series, charge on each capacitor is equal in magnitude, and voltages add. The reciprocal of the equivalent capacitance equals the sum of reciprocals: 1/C_eq = 1/C_1 + 1/C_2 + … Because adding positive reciprocals yields a larger denominator, C_eq must be less than the smallest individual capacitor. This mirrors the dual behavior of resistors (resistors in parallel are less than the smallest resistor).

Step-by-Step Solution:
Write the series formula: 1/C_eq = Σ (1/C_i).Observe that Σ (1/C_i) > 1/C_min for any set with more than one element.Therefore 1/C_eq > 1/C_min → C_eq < C_min.Conclude: C_eq is less than the smallest capacitor value in the series chain.

Verification / Alternative check:
Example: C_1 = 4 µF, C_2 = 6 µF → 1/C_eq = 1/4 + 1/6 = 5/12 → C_eq = 12/5 = 2.4 µF, which is less than 4 µF (the smaller of the two), confirming the rule.

Why Other Options Are Wrong:
Option A mixes capacitance with reactance and is dimensionally incorrect. Option B describes parallel capacitors, not series. Option C is a general definition of impedance from Ohm’s law and does not specifically characterize series capacitance. “None” is wrong because a correct statement exists.

Common Pitfalls:
Forgetting to include voltage balancing resistors for high-voltage series stacks; ignoring tolerance and leakage that cause unequal voltage sharing across series capacitors.

Final Answer:
less than the value of the smallest capacitor