When two capacitors are connected in series and connected across 4kV line, the energy stored in the system is 8J. The same capacitors, if connected in parallel across the same line, the energy stored is 36J. Find the individual capacitance ?

1 Answer
Feb 20, 2018

Let,the individual capacitance of the two capacitors were, #C_1# & #C_2#

So, in series,equivalent capacitance will be #(C_1×C_2)/(C_1 + C_2)#

And,as we know,energy stored on connecting a capacitor of capacitance #C# across a voltage difference of #V# is # 1/2 CV^2#

So,from the given information we can write, # 1/2 (C_1×C_2)/(C_1 + C_2) × V^2 =8#

Again,in parallel,equivalent capacitance will be #C_1 + C_2 #

So,we can say,this time energy stored = # 1/2 (C_1+C_2)×V^2 = 36#

Given, #V = 4×10^3#

So,solving both the equations we get, #C_1 =0.012 F# and #C_2=0.006F#