Find the current l through then `10 Omega ` resistance in the network shown in figure.

Explanation

Across A and B two cells are used in parallel so above circuit can be rewdrawn as
`epsi_(eq) = (epsi_(1) r_(2) + epsi_(2) r_(1))/(r_(1) + r_(2)) = (10 xx 1 + 5 xx 2)/(1+2) = (30)/(3)V`
`r_(eq) = ( r_(1) r_(2))/(r_(1) + r_(2)) = (1xx2)/(1+2) = (2)/(3) Omega`
Current through R
` l= (epsi_(eq))/(R+ r_(eq)) = ((20)/(3))/(10+(2)/(3)) = (20)/(32) A = (5)/(8) A`