Series capacitors voltage division: In a purely series capacitive chain, does the smallest capacitance experience the largest voltage drop for a given charge?

Introduction / Context:Voltage division across series capacitors follows the inverse relation to capacitance values. Because series capacitors carry the same charge, the voltage across each is determined by V_i = Q / C_i. This principle is essential when designing capacitive dividers or placing series capacitors across high-voltage rails for rating purposes.

Given Data / Assumptions:

• Ideal series connection, same current and hence same charge magnitude on each element.
• Negligible leakage and identical steady-state charge distribution.
• Sinusoidal or DC conditions where reactive steady state is reached.

Concept / Approach:In series, Q is common to all capacitors. Therefore V_i = Q / C_i: a smaller C produces a larger V_i. The total applied voltage divides inversely with capacitance values. Designers often include voltage-balancing resistors in high-voltage stacks to mitigate imbalance due to leakage and tolerances.

Step-by-Step Solution:

Verification / Alternative check:AC phasor analysis also shows larger magnitude across the smaller capacitance because Xc = 1 / (2 * pi * f * C) is larger, yielding a larger share of the phasor drop.

Why Other Options Are Wrong:

Common Pitfalls:Ignoring leakage and tolerance which can distort division in practice; forgetting to add balancing networks in HV applications.

Final Answer:Correct