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.