Question

Three capacitors of 0.002, 0.004, and 0.006μF capacitance are connected in series. Can 11,000 V be applied to this combination? What will be the voltage across each capacitor? The breakdown voltage of each capacitor is 4000 V. 

Answer 1

Answer: It cannot; the capacitors will break down. The voltage across the capacitors will be: V₁ = 6000 V, V₂ = 3000 V, and V₃ = 2000 V.

Explanation:

From the equality of charges on the capacitors it follows that

V₁C₁ = V₂C₂; V₂C₂ = V₃C₃.

and V₁ + V₂ + V₃ = V.

The result is obtained by solving these equations.