Series capacitors voltage division — two-capacitor stack at 18 V DC Two capacitors are in series: C1 = 4.7 µF and C2 = 3.3 µF. With 18 V DC applied across the series, what is the voltage across C1?

Introduction / Context:Voltage division across series capacitors is a common design and troubleshooting task in power supplies, level shifters, and high-voltage stacks. In series, both capacitors carry the same charge magnitude, and voltages divide inversely with capacitance.

Given Data / Assumptions:

• C1 = 4.7 µF, C2 = 3.3 µF.
• Total applied DC voltage V = 18 V.
• Ideal capacitors; steady-state charge sharing (no leakage mismatch considered).

Concept / Approach:In series, the charge magnitude Q on each capacitor is the same. The series equivalent is Ceq = (C1 * C2) / (C1 + C2). After applying V, Q = Ceq * V. Then the voltage across C1 is V1 = Q / C1, and across C2 is V2 = Q / C2; note V1 : V2 = C2 : C1.

Step-by-Step Solution:Ceq = (C1 * C2) / (C1 + C2) = (4.7 * 3.3) / (4.7 + 3.3) µF = 15.51 / 8 µF ≈ 1.93875 µF.Q = Ceq * V = 1.93875e-6 * 18 ≈ 34.8975e-6 coulomb.V1 = Q / C1 = 34.8975e-6 / 4.7e-6 ≈ 7.425 V.Rounded to one decimal place, V1 ≈ 7.4 V.

Verification / Alternative check:Use ratio method: V1 : V2 = C2 : C1 = 3.3 : 4.7. Total V = 18 V so V1 = 18 * (3.3 / (4.7 + 3.3)) = 18 * (3.3 / 8) = 18 * 0.4125 ≈ 7.425 V. Same result.

Why Other Options Are Wrong:

• 3.3 V and 6.6 V: underestimate the share on the larger-value capacitor; division is inverse with capacitance.
• 9.4 V: closer to the other branch; here C1 is larger, so it gets less voltage than C2, not more.

Common Pitfalls:

• Dividing voltage directly in proportion to capacitance (should be inverse).
• Ignoring tolerance or leakage matching in real designs, which can unbalance voltages.

Final Answer:7.4 V