Three capacitors of capacitances 3 μ F,9 μ F and 18 μ F are connected once in series and another time in parallels. The ratio of equivalent capacitances in the two cases (Cs/Cp) will be

Question

Three capacitors of capacitances 3 μ F,9 μ F and 18 μ F are connected once in series and another time in parallels. The ratio of equivalent capacitances in the two cases (Cs/Cp) will be (A) 1: 15 (B) 15: 1 (C) 1: 1 (D) 1: 3

Tardigrade Admin · Accepted Answer

(1/Cs) = (1/3) + (1/9) + (1/18) = (1/2) . Hence Cs = 2 μ F Cp = 3 + 9 + 18 = 30 μ F Therefore Cs /Cp = 2/30 = 1 /15.

Q. Three capacitors of capacitances $3 μ F$ , $9 μ F$ and $18 μ F$ are connected once in series and another time in parallels. The ratio of equivalent capacitances in the two cases $(C_{s} / C_{p})$ will be

1 : 15

15 : 1

1 : 1

1 : 3

$\frac{1}{C _{s}} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} = \frac{1}{2}$ . Hence $C_{s} = 2 μ F$
$C_{p} = 3 + 9 + 18 = 30 μ F$
Therefore $C_{s} / C_{p}$ = 2/30 = 1 /15.

Q. Three capacitors of capacitances 3μF,9μF and 18μF are connected once in series and another time in parallels. The ratio of equivalent capacitances in the two cases (Cs​/Cp​) will be