Question

An electric circuit requires a total capacitance of 2μF across a potential of 1000 V. Large number of 1μF capacitances are available each of which would breakdown if the potential is more then 350 V. How many capacitances are required to make the circuit ? 

(A) 24 

(B) 12 

(C) 20 

(D) 18

Answer 1

The correct option (D) 18

Explanation:

Let N be no of capacitors in each row.

∴ N = [(required potential)/(potential of each capacitor)]

= [(100)/(350)] = 2.8 = 3

i.e. 3 capacitors are needed in each row, which will be in series connection.

 capacitance of each capacitor = 1μF

∴ capacitance of each row = [(1/1) + (1/1) + (1/1)]^–1 = (1/3)μF

Let there be M such rows, which are in parallel

∴ Total capacitance = (M/3)μF

but given is total capacitance should be 2μF

∴ (M/3) = 2

∴ M = 6

∴ 6 rows with 3 capacitors in each row. hence

Total capacitors = 3 × 6 = 18