# Question 3...............................................

Samyak Jain
333 Points
3 years ago
Energy stored in a capacitor of capacitance C with potential differnce V across it is (1/2)CV2.
Let C1 and C2 be the capacitances. V = 600 V.
Equivalent capacitance when they are in parallel is C1 + C2. So, energy stored in this combination is
(1/2)(C1 + C2)V2 = 18 J   $\dpi{100} \Rightarrow$  (1/2)(C1 + C2)(6000 V)2 = 18 J
Solving, we get C1 + C2 = 10 – 6 F = 1 $\dpi{100} \mu$F
Equivalent capacitance of C1 and C2 when they are in series is C1C2 / (C1 + C2).
So, energy stored is (1/2){C1C2 / (C1 + C2)}V2 = 4 J
(1/2){C1C2 / (C1 + C2)}(6000 V)2 = 4 J,    here put C1 + C2 = 10– 6 F.
Solving we get C1C2 = (2/9) x 10 – 12 F2  =  (2/9) ($\dpi{100} \mu$F)2 .
Thus, we have C1 + C2 = 1 $\dpi{100} \mu$F  and  C1C2 = (2/9) ($\dpi{100} \mu$F)2
i.e. we want factors of 2/9 whose sum is 1. Such factors are 1/3 and 2/3.
$\dpi{100} \therefore$ C1 = 1/3 $\dpi{100} \mu$F  and  C2 = 2/3 $\dpi{100} \mu$F.