At a node, two currents of $5\ \text{A}$ and $3\ \text{A}$ enter, while one current of $4\ \text{A}$ leaves. What must be the value and direction of the fourth branch current?

By KCL, total entering current equals total leaving current. Entering current is $5+3=8\ \text{A}$. One branch already carries $4\ \text{A}$ away, so the remaining branch must account for the balance on the opposite side of the equation. Thus the fourth branch must be $4\ \text{A}$ entering so that the junction can be described consistently as $5+3=4+4$.