This preview shows pages 1–3. Sign up to view the full content.
DirectCurrent Circuits
263
EXECUTE:
Using
2
/,
PV R
=
()
22
111
2 22
28.0 V
28.0 V
/
490 W,
/
327 W, and
1.60
2.40
PVR
==
=
=
ΩΩ
2
2
333
28.0 V
/
163 W
4.80
=
Ω
EVALUATE:
The total power dissipated is
out
1
2
3
980 W.
PP
P
P
=++=
This is the same as the power
in
2.80 V 35.0 A
980 W
PI
=
E
delivered by the battery.
(f)
2
/.
=
The resistors in parallel each have the same voltage, so the power
P
is largest for the one with the
least resistance.
26.9.
IDENTIFY:
For a series network, the current is the same in each resistor and the sum of voltages for each resistor
equals the battery voltage. The equivalent resistance is
eq
1
2
3
RR
R
R
=
++
.
2
PIR
=
.
SET UP:
Let
1
1.60
R
=Ω
,
2
2.40
R
,
3
4.80
R
=
Ω
.
EXECUTE:
(a)
eq
1.60
2.40
4.80
8.80
R
=
Ω+
Ω=
Ω
(b)
eq
28.0 V
3.18 A
8.80
V
I
R
=
Ω
(c)
3.18 A
I
=
, the same as for each resistor.
(d)
11
(3.18 A)(1.60
)
5.09 V
VI
R
Ω
=
.
(3.18 A)(2.40
)
7.63 V
R
Ω
=
.
33
(3.18 A)(4.80
) 15.3 V
R
Ω
=
. Note that
123
28.0 V
VVV
++=
.
(e)
(3.18 A) (1.60
) 16.2 W
R
Ω
=
.
(3.18 A) (2.40
)
24.3 W
R
Ω
=
.
(3.18 A) (4.80
)
48.5 W
R
Ω
=
.
Since
2
=
and the current is the same for each resistor, the resistor with the greatest
R
dissipates the
greatest power.
EVALUATE:
When resistors are connected in parallel, the resistor with the smallest
R
dissipates the greatest power.
26.10.
(a)
IDENTIFY:
The current, and hence the power, depends on the potential difference across the resistor.
SET UP:
/
2
=
EXECUTE:
(a)
(5.0 W)(15,000
)
274 V
VP
R
Ω
=
(b)
/
(120 V) /(9,000
)
1.6 W
Ω
=
SET UP:
(c)
If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe.
Therefore the maximum power in the larger resistor must be 2.00 W. Use
2
=
to find the maximum current
through the series combination and use Ohm’s law to find the potential difference across the combination.
EXECUTE:
2
=
gives
I
=
P
/
R
= (2.00 W)/(150
Ω
) = 0.0133 A. The same current flows through both
resistors, and their equivalent resistance is 250
Ω
. Ohm’s law gives
V
=
IR
= (0.0133 A)(250
Ω
) = 3.33 V.
Therefore
P
150
= 2.00 W and
100
R
2
=
= (0.0133 A)
2
(100
Ω
) = 0.0177 W.
EVALUATE:
If the resistors in a series combination all have the same power rating, it is the
largest
resistance that
limits the amount of current.
26.11.
IDENTIFY:
For resistors in parallel, the voltages are the same and the currents add.
eq
1
2
R
=+
so
12
eq
,
R
=
+
For resistors in series, the currents are the same and the voltages add.
eq
1
2
R
=
+
.
SET UP:
The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits
shown in Figure 26.11.
EXECUTE:
eq
5.00
R
. In Figure 26.11c,
60.0 V
12.0 A
5.00
I
Ω
. This is the current through each of the
resistors in Figure 26.11b.
12
12
(12.0 A)(2.00
)
24.0 V
R
Ω
=
.
34
34
(12.0 A)(3.00
)
36.0 V
R
Ω
=
. Note
that
12
34
60.0 V
VV
+=
.
12
V
is the voltage across
1
R
and across
2
R
, so
12
1
1
24.0 V
8.00 A
3.00
V
I
R
=
Ω
and
12
2
2
24.0 V
4.00 A
6.00
V
I
R
=
Ω
.
34
V
is the voltage across
3
R
and across
4
R
, so
34
3
3
36.0 V
3.00 A
12.0
V
I
R
=
Ω
and
34
4
4
36.0 V
9.00 A
4.00
V
I
R
=
Ω
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document264
Chapter 26
EVALUATE:
Note that
12 34
II II
+=+
.
Figure 26.11
26.12.
IDENTIFY:
Replace the series combinations of resistors by their equivalents. In the resulting parallel network the
battery voltage is the voltage across each resistor.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Current, Power

Click to edit the document details