In the given cirucit, determine current through branch having indicated resistor `R_(4)` Given `R_(1) = R_(2) = 4Omega, R_(3) = R_(4) = 2Omega`
`R_(5) = 5Omega epsilon_(1) =5 V, epsilon _(2) =10V`

Explanation

Assign potential at each node as shown in figure.

At node 'e', `I_(2) = I_(3)+I_(4)`
where `I_(3) = ((x+5)-x)/(2)` and `I_(4) = ((x+5)-10)/(2)`
Hence, `I_(2) = ((x+5)-x)/(2)+((x+5)-10)/(2)` (i)
Note the outgoing current at node c is equal to incoming current
at node b. Thus our equation is at node C and node b.
`((x+5))/(2)+((0-x))/(4) = I_(2) +((x -0))/(4)` (ii)
From Eqs. (i) and (ii), we get `x = -2.5V`
current through `R_(4) is 3.75A`.