The minimum number of 8 μ F and 250 V capacitors which are used to make a combination of capacitance 16 μ F and voltage 1000 V is

Step-by-step solution

Let m rows of n series capacitor be taken then minimum number of capacitors required is Also effective voltage is V ′ ′ = 1000 = n × 250 ⇒ n = 1000 250 = 4 Also these four capacitor are connected in series then effective capacitance is 1 C ′ ′ = 1 8 + 1 8 + 1 8 + 1 8 = 4 8 ⇒ C ′ ′ = 2 μ F ∴ C ′ ′ ′ ′ = 16 = 2 × m ⇒ m = 16 2 = 8 Hence N = m × n = 8 × 4 = 32