This preview shows page 1. Sign up to view the full content.
111. (a) Placing a wire (of resistance
r
) with current
i
running directly from point
a
to
point
b
in Fig. 2761 divides the top of the picture into a left and a right triangle. If we
label the currents through each resistor with the corresponding subscripts (for instance,
i
s
goes toward the lower right through
R
s
and
i
x
goes toward the upper right through
R
x
),
then the currents must be related as follows:
01
1
2
20
,
,
s
sx
x
ii
i
i
iiiiii
=
+=
+
+
=+
=
where the last relation is not independent of the previous three. The loop equations for the
two triangles and also for the bottom loop (containing the battery and point
b
) lead to
iR iR ir
iR iR iR
ss
x
xx
−
−
=
−−
=
−−−
=
11
2
00
0
0
0
ε
.
We incorporate the current relations from above into these loop equations in order to
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Current, Resistance

Click to edit the document details