Assignment 5 Solutions
Chpt 26
9.
We use
v
d
= J
/
ne = i
/
Ane
. Thus,
()
( ) ( )( )
14
2
28
3
19
2
0.85m
0.21 10
m
8.47 10 / m
1.60 10
C
/
300A
8.1 10 s
13min.
d
LLL
A
n
e
t
vi
A
n
e
i
−−
×××
==
=
=
=×
=
Chpt 27
12. (a) We solve
i
= (
ε
2
–
1
)/(
r
1
+
r
2
+
R
) for
R
:
R
i
rr
=
−
−− =
−
×
−−=
×
−
21
12
3
2
30
20
99 10
..
.
VV
1.0 10
A
ΩΩ
.
Ω
(b)
P = i
2
R
= (1.0
×
10
–3
A)
2
(9.9
×
10
2
Ω
) = 9.9
×
10
–4
W.
26. (a)
R
eq
(
FH
) = (10.0
Ω
)(10.0
Ω
)(5.00
Ω
)/[(10.0
Ω
)(10.0
Ω
) + 2(10.0
Ω
)(5.00
Ω
)] =
2.50
Ω
.
(b)
R
eq
(
FG
) = (5.00
Ω
)
R
/(
R
+ 5.00
Ω
), where
R
= 5.00
Ω
+ (5.00
Ω
)(10.0
Ω
)/(5.00
Ω
+ 10.0
Ω
) = 8.33
Ω
.
So
R
eq
(
FG
) = (5.00
Ω
)(8.33
Ω
)/(5.00
Ω
+ 8.33
Ω
) = 3.13
Ω
.
32. Using the junction rule (
i
3
=
i
1
+
i
2
) we write two loop rule equations:
10.0 V –
i
1
R
1
– (
i
1
+
i
2
)
R
3
= 0
5.00 V –
i
2
R
2
– (
i
1
+
i
2
)
R
3
= 0.
(a) Solving, we find
i
2
= 0, and
(b)
i
3
=
i
1
+
i
2
= 1.25 A (downward, as was assumed in writing the equations as we did).
41. (a) The batteries are identical and, because they are connected in parallel, the
potential differences across them are the same. This means the currents in them are the
1
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i
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 Spring '08
 Parsons
 Resistor, loop rule, RC dt, junction rule, loop rule equations

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