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.