Question

An electrical technician requires a capacitance of 2μF in a circuit across a potential difference of 1KV. A large number of 1μF capacitors are available to him, each of which can withstand a potential difference of not than 400V. suggest a possible arrangement that requires a minimum number of capacitors.

(A) 2 rows with 2 capacitors

(B) 4 rows with 2 capacitors

(C) 3 rows with 4 capacitors

(D) 6 rows with 3 capacitors

Answer 1

The correct option (D) 6 rows with 3 capacitors

Explanation:

Total capacitance = C = 2μF

V = potential difference = 100 V

C1 = capacitance of each capacitor = 1μF

V1 = maximum potential difference = 400 V

Let n capacitors of 1μF each be connected in series in a row and m such rows be in parallel as shown.

∴ Potential difference across each capacitor = [(1000)/n] = 400

∴ n = 2.5

∴ n = 3

capacitance of each row of 3 capacitors of 1μF = (1/3)μF

∴ capacitance of m such rows in parallel = (m/3)μF

∴ (m/3)μF = 2μF

∴ m = 6

∴ 6 rows with 3 capacitors in each row.