In one part of above network, steady current is flowing. Values of resistances are as shown in the diagram. Find energy stored in capacitor.

Explanation

When the capacitor becomes fully charged, there is no current in the branch containing capacitor. Now, current in the branch AD and DE is 1 + 2 = 3 A.
Current in the branch EB is 1 A.
Now, `V_(AD) = 3 xx 5 = 15 ` V
`V_(DE) = 3 xx 1 = 3 V `
` V_(EB) = 1 xx 2 = 2 `V
Potential difference across capacitor is,
`V = V_(AD) + V_(DE) + V_(EB)`
= 15 + 3 + 2
`therefore V = 20 ` Volt
Now, electrostatic potential energy stored in the capacitor is,
`U_(E) = (1)/(2) CV^(2)`
`therefore U_(E) = (1)/(2) xx 4 xx 10^(-6) xx (20)^(2)`
`therefore U_(E) = 8 xx 10^(-4) `J