P28_096.PDF

96. (Fourth problem of Cluster )

(a) The symmetry of the problem allows us to use i 2 as the current in both of the R 2 resistors and i 1 for

the R 1 resistors. We see from the junction rule that i 3 = i 1 −

i 2 . There are only two independent

loop rule equations:

E−

i 2 R 2 −

i 1 R 1 =0

E−

2 i 1 R 1 −

( i 1 −

i 2 ) R 3 =0 .

where in the latter equation, a zigzag path through the bridge has been taken. Solving, we ﬁnd

i 1 =0 . 002625 A , i 2 =0 . 00225 A and i 3 = i 1 −

i 2 =0 . 000375 A. Therefore, V A −

V B = i 1 R 1 =5 . 25 V.

(b) It follows also that V B −

V C = i 3 R 3 =1 . 50V.

V D = i 1 R 1 =5 . 25 V.

(d) Finally, V A −

V C = i 2 R 2 =6 . 75 V.