From the given currents in the diagram, the current through the middle branch of the circuit must be 1.00 A (the difference between 2.00 A and 1 .00 A). Using Kirchhoff's rules and passing counterclockwise around the top loop, we have
20−1×1+4×1+1×1−ε1−6×1=0orε1=18V
Now traveling around the external loop of the circuit, we have
20−1×1−1×2−ε2−2×2−6×1=0orε2=7V
and Va+4×1+1×1−ε1=VborVb−Va=5−18=−13V