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 redrawn as
`epsilon_(eq) = ( epsilon_(1)r_(2)+epsilon_(2)r_(1))/(r_(1)+r_(2))= ( 10 xx 1 + 5 xx 2 )/( 1 + 2) = ( 20)/(3)V`
`r_(eq) = (r_(1)r_(2))/(r_(1)+r_(2))= ( 1 xx 2 )/( 1+ 2 ) = ( 2)/(3) Omega `

Current through R
`l = ( epsilon_(eq))/(R + r_(eq)) = ( ( 20)/(3))/(10+(2)/(3))= (20)/(32) A = (5)/(8) A`