Solution
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−ε
1
−6×1=0orε
1
=18V
Now traveling around the external loop of the circuit, we have
20−1×1−1×2−ε
2
−2×2−6×1=0orε
2
=7V
and V
a
+4×1+1×1−ε
1
=V
b
orV
b
−V
a
=5−18=−13V