(a) Find the current in the `20(Omega)`resistor shown in figure.(b)If a capacitor of capacitance `4(mu)F is joined between the point A and B,what would be the electrostatic energy stored in it in steady state?

Explanation

Taking circuit abeda,
` 10i + 20(i-i_1) - 5 = 0 `
` rArr 10i + 20i - 20i -5 =0 `
` rArr 30i - 20i_1 -5 = 0 `
Taking circuit abfca,
` 20(i-i_1)- 5 - 10i_1 = 0 `
` rArr 20i -10i_1 - 20i_1 - 5 =0 `
` rArr 20i - 30i_1 -5 =0 `
` (30i-20i_1 -5 =0 )2 `
` (20i - 30i_1-5 =0 )3 `
` (90-40)i_1 =0 `
` rArr i_1 = 0 `
`rArr 30i - 5 = 0 `
` i = 5/30 = 0.16A`
` Current through 20 Omega is 0.16A. `