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EEJ MAIN Physics QUESTION #6988
Question 1

A capacitance of 2 μF is needed across a potential difference of 1.0 kV. Only 1 μF capacitors rated for a maximum of 300 V each are available. What is the minimum number of such capacitors required?

  • 32✔️
  • 2
  • 16
  • 24
Correct Answer Explanation

Step 1: Series groups to handle voltage

Each capacitor handles 300 V. For 1000 V: minimum capacitors in series = $\lceil 1000/300 \rceil = 4$ (since $3 \times 300 = 900 < 1000$, we need 4 in series).

4 capacitors in series gives capacitance $= \frac{1}{4}\ \mu\text{F}$ per row.

Step 2: Parallel rows to achieve 2 μF

Number of rows in parallel $= \frac{2}{1/4} = 8$ rows.

Total capacitors $= 4 \times 8 = \mathbf{32}$