Difficulty: Easy
Correct Answer: 4 µF
Explanation:
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:
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:
Common Pitfalls:
Final Answer:
4 µF
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