Minimum number of capacitors of 2 µF each required to obtain a capa...

Question

Minimum number of capacitors of 2 µF each required to obtain a capacitance of 5 µF will be

Accepted Answer

✅ Correct Answer: (A) 4

Arrangement Strategy:

Step 1 — 3 capacitors in parallel:

C_parallel = 2 + 2 + 2 = 6 µF

Step 2 — 1 capacitor in series with above:

1/C_eq = 1/6 + 1/2 = 1/6 + 3/6 = 4/6

C_eq = 6/4 = 1.5 µF ✗ (Not 5 µF)

Correct Arrangement — 2 in parallel + 2 in series grouping:

Group 1: 2 capacitors in series → C = 1 µF
- Group 2: 2 capacitors in parallel → C = 4 µF
  - Combine Group 1 + Group 2 in parallel → C = 1 + 4 = 5 µF ✓

∴ <b>Minimum 4 capacitors</b> are needed.