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 − 2i
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.
(c) We find V
C
− V
D
= i
1
R
1
= 5.25 V.
(d) Finally, V
A
− V
C
= i
2
R
2
= 6.75 V.