Two capacitors of 0.005 μF, 40 V and 0.02 μF, 100 V are connected in series. What is the effective capacitance of the series combination and the maximum DC voltage that can be applied across it?

Concept:

When capacitors are connected in series, the total (effective) capacitance is given by:

\( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \)

The maximum voltage across the combination is limited by the capacitor which reaches its voltage rating first, considering how voltage divides in series based on inverse of capacitance.

Given:

• C₁ = 0.005 µF = 5 nF, V_max1 = 40 V
• C₂ = 0.02 µF = 20 nF, V_max2 = 100 V

Step 1: Effective Capacitance

\( \frac{1}{C_{eq}} = \frac{1}{5} + \frac{1}{20} = \frac{4 + 1}{20} = \frac{5}{20} = \frac{1}{4} \Rightarrow C_{eq} = 4~nF \)

Step 2: Maximum DC Voltage

In series, same charge appears across both capacitors. Let charge Q be the same:

\(Q = C_1 V_1 = C_2 V_2 \Rightarrow \frac{V_1}{V_2} = \frac{C_2}{C_1} = \frac{20}{5} = 4 \)

Let V₂ = x, then V₁ = 4x ⇒ Total voltage V_total = V₁ + V₂ = 4x + x = 5x

Now, apply voltage limit for C₁: V₁ ≤ 40 V ⇒ 4x ≤ 40 ⇒ x ≤ 10 V

⇒ Total V_total = 5x = 50 V (Maximum allowable without exceeding capacitor voltage ratings)

Hence,  the correct answer is option 2