Question

Question 5 1 pts 5) Three capacitors, all initially without any dielectrics inserted, of C1 4 micro- farads, C2 7 micro-farads, and C3 12 micro-farads, are connected purely in series to a battery and the total stored energy of all three capacitors is 248 micro-joules. Then, the plates of C1 are brought closer by a factor of 4 and a dielectric with a dielectric constant of 1.6 is inserted into C2 while still connected in series to the same battery. (C3 does not change at all.) How much does the total stored energy of all three capacitors increase in micro-joules after the capacitors are changed? Not the factor change... how much does the stored energy go up in value.

Answer 1

We know that, equivalent capacitor when connected in series is,

C C1 C2 C3

C 4712

C=2.1 micro farady.

And energy stored is,

E=0.5CV²

248*10^-6=0.5*2.1*10^-6*V²

V=15.368 V

C1 will increase by a factor of 4.

C1=16,

And C2=7*1.6

=11.2

C3=12,

New equivalent capacitor is,

C 16 11.212

C=4.253 micro farady.

Now new energy,

E=0.5CV²

E=0.5*4.253*15.368²

E=502.259 micro joule

So the increase in the value will be,

=502.259-248

=254.259