Question

Minimum numbers of 8 µF and 250 V capacitors

used to make a combination of 16 µF and 1000 V

are:

(1) 4 (2) 32

(3) 8 (4) 3​

Answer 1

Answer:

Correct Option is :

(2). 32

Explanation:

To hold a potential of 1 kV or 1000 V, the minimum number of capacitors required in each series combination = $\mathsf{\dfrac{1000 V} {250 V} = 4 }$

Equivalent capacitance of such a series combination is

$\mathsf{\dfrac{1} {C_s}} = {\dfrac{1}{1}} + {\dfrac{1}{1}} + {\dfrac{1}{1}} + {\dfrac{1}{1}}$

$\mathsf{{C_s} = {\dfrac{1}{4}}\mu{F}}$

For $\mathsf{C_{eq}}$ to be 2μF, 8 such parallel combinations are required.

Minimum number of capacitors needed = 4 × 8 = 32