In the figure Figure 1 find the energy stored in the 10F ca

In the figure (Figure 1) , find the energy stored in the 1.0-?F capacitor when a 51 V battery is connected between points A and B.

Solution

Two Capactiors are in parallel = (2+1) uF = 3uF

Now,
3.0 uF, 3 uF & 2 uF are in series

Cnet = 1/(1/3 + 1/3 + 1/2 ) uF
Cnet = 0.86 uF

Now, Q = C*V
Q = 0.86 * 51 uC
Q = 43.86 uC

Charge on 1.0 uF Capacitor = 1/(1+2) * 43.86 uC = 21.93 uC

Energy stored in 1.0-F capacitor = Q^2/2*C
Energy stored in 1.0-F capacitor = (21.93*10^-6)^2/ (2*1*10^-6) J
Energy stored in 1.0-F capacitor = 2.4 * 10^-4 J