Find the energy stored in this capacitance

Question

Two capacitors, c1= 18.0 uF and c2=36.0 uF are connected in series and a 12.0-V battery is connected across the two capacitors. 

 

a) find the equivalent capacitance 

b) find the energy stored in this equivalent capacitance

c) find the energy stored in each individual capacitor 

d) show that the sum of these two energies is the same as the energy found in part b

e) will this equality always be true or does it depend on the number of capacitors and their capacitances?

f) if the same capacitors were connected in parallel, what potential difference would be required across them so that the    combination stores the same energy as in part a

g) which capacitor stores more energy in this situation, c1 or c2?

Answer 1

a-------

1/c=(1/18)+(1/36)

 

c=12 micron

 

b----------

Energy stored= (1/2)*c *v^2.   joules

 

            =0.5*12*(12^2)

             =0.864*10^(-3) joules

 c-------

           q = cv

              =12*12=144

      voltage across 18 microf cap=

       v =q/c

 

         =144/18=8v

Energy stored in this cap = 0.5*18*8^2=0.576*10^(-3) joules

 

      voltage across 36 v

   v= 144/36=4v

 

Energy stored in this cap =0.5*36*4^2

 

                                     =0.288*10*(-3)  joules

d -------

Total energy = (0.576+0.288)*10^(-3)

                    =0.864*10^(-3) joules

 

e-------

Always true

 

f--------

  c = 18+36=54 micron

 

I hope you can now do the rest yourself.