Series combination of capacitors: given 2 µF, 4 µF, and 10 µF in series, the total capacitance is less than which value?

Introduction / Context:
Capacitors in series behave differently from resistors: the total capacitance decreases and is always less than the smallest individual capacitor. Knowing this qualitative rule helps catch design and exam errors quickly, even before calculating exact values.

Given Data / Assumptions:

• Three capacitors: 2 µF, 4 µF, and 10 µF connected in series.
• Ideal capacitors; no leakage or parasitics considered.

Concept / Approach:
The series capacitance relation is 1/C_total = 1/C1 + 1/C2 + 1/C3. Since each term is positive, 1/C_total > 1/C_smallest, which implies C_total < C_smallest. Thus the total must be less than 2 µF, the smallest of the three.

Step-by-Step Solution:

Verification / Alternative check:
If desired, compute: 1/C_total = 1/2 + 1/4 + 1/10 = 0.5 + 0.25 + 0.1 = 0.85 µF^-1, so C_total ≈ 1 / 0.85 ≈ 1.176 µF, which is indeed less than 2 µF.

Why Other Options Are Wrong:

• 4 µF, 10 µF: These are larger than the smallest value; the total cannot exceed the smallest in a series network.
• 1.5 µF: This is a candidate numeric value but the prompt asks “less than which value?” The correct qualitative bound is the smallest capacitor, 2 µF.

Common Pitfalls:

• Applying resistor intuition (adding values in series) to capacitors; capacitors in series do the opposite.
• Arithmetic mistakes with reciprocals; the inequality method is a fast sanity check.

Final Answer:
2 µF