Series capacitors charged from 24 V: a 2 µF capacitor has 16 V across it and the other series capacitor has 8 V. What is the value of the unknown capacitor?

Introduction / Context:In series, capacitors share the same charge magnitude, and voltages divide inversely with capacitance. This is routinely used in high-voltage stacks and sensor circuits. Here we find an unknown capacitance from measured voltage division across series caps.

Given Data / Assumptions:

• Two series capacitors across 24 V.
• C1 = 2 µF has V1 = 16 V.
• Unknown C2 has V2 = 8 V.
• Ideal capacitors; same charge magnitude on each.

Concept / Approach:For capacitors in series: charge Q is equal on both. Voltage V = Q / C, so the voltage ratio equals the inverse capacitance ratio. Therefore, V1 / V2 = C2 / C1, which can be solved for C2.

Step-by-Step Solution:

Verification / Alternative check:Compute equivalent C_total: 1/C_total = 1/2 + 1/4 = 0.5 + 0.25 = 0.75 ⇒ C_total = 1.333 µF. Series charge Q = C_total * V_total = 1.333e-6 * 24 = 32 µC. Then V1 = Q/C1 = 32e-6 / 2e-6 = 16 V; V2 = 32e-6 / 4e-6 = 8 V, confirming the solution.

Why Other Options Are Wrong:

• 1 µF or 2 µF: Would not yield the given 16 V/8 V split.
• 8 µF: Would give V1/V2 = C2/C1 = 8/2 = 4, implying V1 = 4 * V2, not 2:1 as given.

Common Pitfalls:

• Using resistor-style division (direct proportionality) instead of inverse proportionality for capacitors in series.
• Forgetting that series capacitors carry equal charge.

Final Answer:4 µF