R
a
.
r
a
5
2.63
3
10
2
8
V
#
m.
r
c
5
1.72
3
10
2
8
V
#
m
R
5
r
L
A
.
1
115.2 V
2
/
2
5
57.6 V.
6.0
V
19.2 V
1
96.0 V
5
115.2 V.
20.0
V
4.0
V
4.0
V
1
2.4 A
21
8.0
V
2
5
19.2 V.
8.0
V
16.0
V
16.0
V
8.0
V
18.0
V
9.0
V
12.0
V
6.0
V
6.0
V
24.0
V
8.0
V
4.0
V
20.0
V
6.0
V
20.0
V
2.4 A
x
x
y
x
y
x
y
y
a
(
a
)
(
b
)
(
c
)
(
d
)
a
8.0
V
.
1922
Chapter 19
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Reflect:
The parallel combination has less equivalent resistance even though both cables contain the same volume of
each metal.
19.88.
Set Up:
Because of the polarity of each emf, the current in the
resistor must be in the direction
shown in Figure 19.88. Let
I
be the current in the 24.0 V battery.
Figure 19.88
Solve:
The loop rule applied to loop (1) gives:
The
junction rule then says that the current in the middle branch is 2.00 A, as shown in Figure 19.88b. The loop rule
applied to loop (2) gives:
and
19.89.
Set Up:
There is no current in the middle branch because there is not a complete conducting path for that
branch. There is only a single current in the circuit, as shown in Figure 19.89. To find
start at point
b
and travel to point
a.
Figure 19.89
Solve:
Going from
a
to
b
through the 12.0 V battery gives
Or, going from
a
to
b
through the 8.0 V battery gives
and
V
a
2
V
b
5 2
2.0 V
1
1
0.444 A
21
5.00
V
2
5 1
0.22 V.
V
a
2
I
1
2.00
V
2
2
I
1
1.00
V
2
2
8.0 V
2
I
1
2.00
V
2
1
10.0 V
5
V
b
V
a
2
V
b
5
12.0 V
2
10.0 V
2
1
0.444 A
21
4.00
V
2
5 1
0.22 V.
V
a
1
I
1
2.00
V
2
1
I
1
1.00
V
2
2
12.0 V
1
I
1
1.00
V
2
1
10.0 V
5
V
b
.
I
5
0.444 A.
12.0 V
2
8.0 V
2
I
1
2.00
V 1
2.00
V 1
1.00
V 1
2.00
V 1
1.00
V 1
1.00
V
2
5
0.
I
a
b
I
I
I
12.0 V
10.0 V
1.00 V
8.0 V
1.00
V
1.00
V
3.00
V
2.00
V
2.00
V
2.00
V
1.00
V
+
+
+
V
ab
5
V
a
2
V
b
,
E
5
8.6 V.
1
E
2
1
1.80 A
21
7.00
V
2
1
1
2.00 A
21
2.00
V
2
5
0
I
5
3.80 A.
1
24.0 V
2
1
1.80 A
21
7.00
V
2
2
I
1
3.00
V
2
5
0.
I
24.0 V
24.0 V
3.80 A
2.00 A
1.8 A
1.80 A
7.00
V
2.00
V
3.00
V
7.00
V
2.00
V
3.00
V
1
1
2
1
2
2
(
a
)
(
b
)
+
+
+
+
E
E
7.00
V
Current, Resistance, and DirectCurrent Circuits
1923
The voltmeter reads 0.22 V.
is positive, so point
a
is at higher potential.
Reflect:
Since there is no current in the middle branch, the two resistances in that branch could be removed without
affecting the potential difference
The 10.0 V emf doesn’t affect the current but does affect
19.90.
Set Up:
The charge is reduced to
of its maximum value when
Solve: (a)
At
(b)
19.91.
Set Up:
At
Solve: (a)
At
and
(b)
so
19.92.
Set Up:
Zero current through the galvanometer means the current
through
N
is also the current through
M
and the current
through
P
is the same as the current through
X
. And it means that points
b
and
c
are at the same
potential, so
Solve: (a)
The voltage between points
a
and
d
is
so
and
Using these expressions in
gives
and
(b)
X
5
MP
N
5
1
850.0
V
21
33.48
V
2
15.00
V
5
1897
V
X
5
MP
/
N
.
NX
5
PM
N
1
P
1
X
2
5
P
1
N
1
M
2
.
E
N
1
M
N
5
E
P
1
X
P
.
I
1
N
5
I
2
P
I
2
5
E
P
1
X
.
I
1
5
E
N
1
M
E
,
I
1
N
5
I
2
P
.
I
2
I
1
C
5
t
R
5
6.00 s
5.00
3
10
5
V
5
1.20
3
10
2
5
F.
t 5
RC
R
5
E
i
5
400.0 V
0.800
3
10
2
3
A
5
5.00
3
10
5
V
.
E
5
iR
t
5
0,
t 5
RC
.
q
5
0.
t
5
0,
E
5
iR
1
q
C
.
t
5 t 5
1
4600
V
21
0.800
3
10
2
9
F
2
5
3.68
3
10
2
6
s
5
3.68
m
s.
q
5
iRC
5
1
0.250 A
21
4600
V
21
0.800
3
10
2
9
F
2
5
9.20
3
10
2
7
C
5
0.920
m
C.
t
5
0,
t
5 t 5
RC
.
1
/
e
iR
5
q
C
.
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 Spring '09
 RODRIGUEZ
 Physics, Charge, Current, Resistance, Resistor, Electrical resistance, Series and parallel circuits, DirectCurrent Circuits

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