P28_021.PDF

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 . 0V

100Ω =0 . 050A .

The second yields

i 2 = E 1 −E 2 −E 3

R 2

= 6 . 0V

−

5 . 0V

−

4 . 0V

−

0 . 060A .

50Ω

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 . 0V+5 . 0V=9 . 0V.