The circuit shown in fig. Is in steady state.

i. Find the energy stored in the capacitors shown in fig.
ii. Find the rate at which battery supplies energy.

Explanation

Since in steady state, no current flows thruogh the capacitors,
the current through `1 Omega` resistor becomes zero. Current throgh
resistor and charge on capacitors will be as shown in fig.
Applying KVL on mesh MACDA, we get
`2I +3I +3I +2I - 10 =0 OR I = 1A`
Mesh MABM: ` 10 - 2I - (q_(1))/(2xx10^(-6)) = 0 or q_(1) = 16muC`
Mesh MBDM: `-(q_(2))/(2xx10^(-6))-2I=0 or q_(2)= -4muC`
Mesh MDCNM: `2I+3I-(q_(3))/((1xx10^(-6))) or q_(3)=5muC`
Energy stored in capacitors is
`U =sum(q^(2))/(2C)=(q_(1)^(2))/(2xx(2xx10^(-6)))+(q_2^2)/(2xx(2xx10^(-6)))+( q_3^2)/(2xx(1xx10^(-6)))`
`=80.5xx10^(-6)J`
Rate of supply of energy is `P = EI = 10xx1 = 10W`