21. Let i
1
be the current in R
1
and take it to be positive if it is to the right. Let i
2
be the current in R
2
and take it to be positive if it is upward. When the loop rule is applied to the lower loop, the result is
E
2
− i
1
R
1
= 0 .
and when it is applied to the upper loop, the result is
E
1
− E
2
− E
3
− i
2
R
2
= 0 .
The first equation yields
i
1
=
E
2
R
1
=
5.0 V
100 Ω
= 0.050 A .
The second yields
i
2
=
E
1
− E
2
− E
3
R
2
=
6.0 V
− 5.0 V − 4.0 V
50 Ω
=
−0.060 A .
The negative sign indicates that the current in R
2
is actuallydownward. If V
b
is the potential at point
b, then the potential at point a is V
a
= V
b
+
E
3
+
E
2
, so V
a
− V
b
=
E
3
+
E
2
= 4.0 V + 5.0 V = 9.0 V.