Physics 2213
HW #5 – Solutions
Spring 2010
26.23. [Kirchoff’s Circuit Rules]
We have to use Kirchoff’s rules to find the
unknown currents, emfs, and resistances.
Apply the junction rule at points
a, b, c
and
d
to calculate the unknown currents. Then
apply the loop rule to three loops to
calculate
1
2
,
and .
R
ε
ε
The circuit is sketched
to the right.
(a)
Apply the junction rule to point
a
:
3
3.00 A
5.00 A
0
I
+
−
=
3
8.00 A
I
=
Apply the junction rule to point
b
:
4
2.00 A
3.00 A
0
I
+
−
=
4
1.00 A
I
=
Apply the junction rule to point
c
:
3
4
5
0
I
I
I
−
−
=
5
3
4
8.00 A
1.00 A
7.00 A
I
I
I
=
−
=
−
=
As a check, apply the junction rule to point
d
:
5
2.00 A
5.00 A
0
I
−
−
=
5
7.00 A
I
=
(b)
Apply the loop rule to loop(1):
(
)(
)
(
)
1
3
3.00 A
4.00
3.00
0
I
ε
−
Ω −
Ω =
(
)(
)
1
12.0 V
8.00 A
3.00
36.0 V
ε
=
+
Ω =
Apply the loop rule to loop (2):
(
)(
)
(
)
2
3
5.00 A
6.00
3.00
0
I
ε
−
Ω −
Ω =
(
)(
)
2
30.0 V
8.00 A
3.00
54.0 V
ε
=
+
Ω =
(c)
Apply the loop rule to loop (3):
(
)
1
2
2.00 A
0
R
ε
ε
−
−
+
=
2
1
54.0 V
36.0 V
9.00
2.00 A
2.00 A
R
ε
ε
−
−
=
=
=
Finally, we can apply the loop rule to loop (4) as a check of our calculations:
(
)
(
)(
)
(
)(
)
2.00 A
3.00 A
4.00
5.00 A
6.00
0
R
−
−
Ω +
Ω =
(
)(
)
2.00 A
9.00
12.0 V
30.0 V
0
−
Ω −
+
=
18.0 V
18.0 V
0
−
+
=
26.50. [Electric Dryer]
(a)
VI
P
=
,
so
I
=
P
V
=
4100 W
240 V
=
17.1 A.
Recall from page 201 that 12-gauge wire can carry up to 20A
safely.
Thus, we can use 12-gauge wire safely for this electric dryer.
(b)
2
V
P
VI
R
=
=
.
Solving for the resistance,
R
=
V
2
P
=
(240 V)
2
4100 W
=
14
Ω
.
(c)
At
11
/
c
per kWH, for 1 hour the cost is
(11
/
c/kWh)(1 h)(4.1 kW)
=
45
/
c
.
(d)
It makes sense that the cost to operate the device is proportional to its power consumption.
The number also sounds about right.
I am charged about one dollar to run a load of clothes
through a dryer for an hour.

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