Question

What total energy is stored in the capacitors in the figure below

(C₁ = 0.643

Answer 1

Here C1 and C2 are in series then eqivalent capacitance is

                1/Ceq =1/C1+1/C2

Then Ceq =C1C2/C1+C2 =0.627uF

Now Ceq and 2.50uF are in parallel then

                       C = 0.627+2.50 =3.127uF

We know that in parallel combination voltage will be the same

                         E =(1/2)CV2 then voltage across the 2.50uF cpacitor is V =Sqrt(2E/C) =12V

Now total energy stored across the capacitors is

                             T.E =(1/2)CV2 =0.5(3.127*10^-6)(12)² =225.144*10^-6J =2.25*10^-4J