Question

Given a number of capacitors labelled as 8μF, 250 V. Find the minimum number of capacitors needed to get an arrangement equivalent to 16 μF, 1000 V

• A. 4
• B. 16
• C. 32
• D. 64

Answer 1

Answer: (C)

32

Answer 2

Let C = 8 μF, C′ = 16 μF and V = 250 volt, V′ = 1000 V

Suppose m rows of given capacitors are connected in parallel which each row contains n capacitor then Potential difference across each capacitors $V = \frac{V'}{n}$ and equivalent capacitance of network $C' = \frac{mC}{n}$.

On putting the values, we get n = 4 and m = 8. Hence total capacitors = m × n = 8 × 4 = 32.

Short Trick : For such type of problem number of capacitors $n = \frac{C'}{C} \times \left( \frac{V'}{V} \right)^{2}$.

Here $n = \frac{16}{8}\left( \frac{1000}{250} \right)^{2} = 32$