2628
Chapter 26
E
XECUTE
:
1
(6.00
3.00
)
36.0 V
0
I
−
Ω +
Ω +
=
1
36.0 V
4.00 A
6.00
3.00
I
=
=
Ω +
Ω
2
(3.00
6.00
)
36.0 V
0
I
−
Ω +
Ω +
=
2
36.0 V
4.00 A
3.00
6.00
I
=
=
Ω +
Ω
To calculate
ab
a
b
V
V
V
=
−
start at point
b
and travel to point
a
, adding up all the potential rises and drops along the
way. We can do this by going from
b
up through the 3.00
Ω
resistor:
2
1
(3.00
)
(6.00
)
b
a
V
I
I
V
+
Ω −
Ω =
(4.00 A)(3.00
)
(4.00 A)(6.00
)
12.0 V
24.0 V
12.0 V
a
b
V
V
−
=
Ω −
Ω =
−
= −
12.0 V
ab
V
= −
(point
a
is 12.0 V lower in potential than point
b
)
E
VALUATE
:
Alternatively, we can go from point
b
down through the 6.00
Ω
resistor.
2
1
(6.00
)
(3.00
)
b
a
V
I
I
V
−
Ω +
Ω =
(4.00 A)(6.00
)
(4.00 A)(3.00
)
24.0 V
12.0 V
12.0 V,
a
b
V
V
−
= −
Ω +
Ω = −
+
= −
which checks.
(b) I
DENTIFY
:
Now there are multiple current paths, as shown in Figure 26.73b. Use junction rule to write the
current in each branch in terms of three unknown currents
I
1
,
I
2
, and
I
3
. Apply the loop rule to three loops to get
three equations for the three unknowns. The target variable is
I
3
, the current through the switch.
R
eq
is calculated
from
eq
,
V
IR
=
where
I
is the total current that passes through the network.
S
ET
U
P
:
The three unknown currents
1
2
3
,
, and
I
I
I
are labeled on Figure 26.73b.
Figure 26.73b
E
XECUTE
:
Apply the loop rule to loops (1), (2), and (3).
loop (1):
1
3
2
(6.00
)
(3.00
)
(3.00
)
0
I
I
I
−
Ω +
Ω +
Ω =
2
1
3
2
I
I
I
=
−
eq.(1)
loop (2):
1
3
2
3
3
(
)(3.00
)
(
)(6.00
)
(3.00
)
0
I
I
I
I
I
−
+
Ω +
−
Ω −
Ω =
2
3
1
2
3
1
6
12
3
0 so 2
4
0
I
I
I
I
I
I
−
−
=
−
−
=
Use eq(1) to replace
2
:
I
1
3
3
1
4
2
4
0
I
I
I
I
−
−
−
=
1
3
1
3
3
6
and
2
I
I
I
I
=
=
eq.(2)
loop (3)
(This loop is completed through the battery [not shown], in the direction from the
to the
−
+
terminal.):
1
1
3
(6.00
)
(
)(3.00
)
36.0 V
0
I
I
I
−
Ω −
+
Ω +
=
1
3
1
3
9
3
36.0 A and 3
12.0 A
I
I
I
I
+
=
+
=
eq.(3)
Use eq.(2) in eq.(3) to replace
1
:
I
3
3
3(2
)
12.0 A
I
I
+
=
3
12.0 A/7
1.71 A
I
=
=
1
3
2
3.42 A
I
I
=
=
2
1
3
2
2(3.42 A)
1.71 A
5.13 A
I
I
I
=
−
=
−
=
The current through the switch is
3
1.71 A.
I
=
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DirectCurrent Circuits
2629
(c)
From the results in part (a) the current through the battery is
1
2
3.42 A
5.13 A
8.55 A.
I
I
I
=
+
=
+
=
The
equivalent circuit is a single resistor that produces the same current through the 36.0 V battery, as shown in
Figure 26.73c.
36.0 V
0
IR
−
+
=
36.0 V
36.0 V
4.21
8.55 A
R
I
=
=
=
Ω
Figure 26.73c
E
VALUATE
:
With the switch open (part a), point
b
is at higher potential than point
a
, so when the switch is closed
the current flows in the direction from
b
to
a
. With the switch closed the circuit cannot be simplified using series
and parallel combinations but there is still an equivalent resistance that represents the network.
26.74.
I
DENTIFY
:
With
S
open and after equilibrium has been reached, no current flows and the voltage across each
capacitor is 18.0 V. When
S
is closed, current
I
flows through the
6.00
Ω
and
3.00
Ω
resistors.
S
ET
U
P
:
With the switch closed,
a
and
b
are at the same potential and the voltage across the
6.00
Ω
resistor
equals the voltage across the
6.00 F
μ
capacitor and the voltage is the same across the
3.00 F
μ
capacitor and
3.00
Ω
resistor.
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 Spring '06
 Buchler
 Physics, Resistor, Electrical resistance, Ω, Series and parallel circuits

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