2510
Chapter 25
25.43.
IDENTIFY:
The bulbs are each connected across a 120V potential difference.
SET UP:
Use
P = V
2
/
R
to solve for
R
and Ohm’s law (
I = V/R
) to find the current.
EXECUTE:
(a)
R = V
2
/
P
= (120 V)
2
/(100 W) = 144
Ω
.
(b)
R = V
2
/
P
= (120 V)
2
/(60 W) = 240
Ω
(c)
For the 100W bulb:
I = V/R
= (120 V)/(144
Ω
) = 0.833 A
For the 60W bulb:
I
= (120 V)/(240
Ω
) = 0.500 A
EVALUATE:
The 60W bulb has
more
resistance than the 100W bulb, so it draws less current.
25.44.
IDENTIFY:
Across 120 V, a 75W bulb dissipates 75 W. Use this fact to find its resistance, and then find the
power the bulb dissipates across 220 V.
SET UP:
P = V
2
/
R
, so
R = V
2
/
P
EXECUTE:
Across 120 V:
R
= (120 V)
2
/(75 W) = 192
Ω
. Across a 220V line, its power will be
P = V
2
/
R
=
(220 V)
2
/(192
Ω
) = 252 W.
EVALUATE:
The bulb dissipates much more power across 220 V, so it would likely blow out at the higher
voltage. An alternative solution to the problem is to take the ratio of the powers.
2
2
2
220
220
220
2
120
120
120
/2
2
0
/1
2
0
PV
R
V
R
V
⎛⎞
==
=
⎜⎟
⎝⎠
.
This gives
2
220
220
(75 W)
120
P
=
= 252 W.
25.45.
IDENTIFY:
A “100W” European bulb dissipates 100 W when used across 220 V.
(a)
SET UP:
Take the ratio of the power in the US to the power in Europe, as in the alternative method for
problem 25.44, using
P = V
2
/
R
.
EXECUTE:
2
2
2
US
US
US
2
EE
E
/
120 V
.
/
220 V
RV
R
V
=
This gives
2
US
120 V
(100 W)
220 V
P
=
= 29.8 W.
(b)
SET UP:
Use
P = IV
to find the current.
EXECUTE:
I = P/V
= (29.8 W)/(120 V) = 0.248 A
EVALUATE:
The bulb draws considerably less power in the U.S., so it would be much dimmer than in Europe.
25.46.
IDENTIFY:
I
=
.
Energy
.
Pt
=
SET UP:
(9.0 V)(0.13 A) 1.17 W
P
EXECUTE:
Energy
(1.17 W)(1.5 h)(3600 s/h)
6320 J
EVALUATE:
The energy consumed is proportional to the voltage, to the current and to the time.
25.47.
IDENTIFY
and
SET UP:
By definition
.
P
p
LA
=
Use
,
and
IEV
L
I J
A
=
to rewrite this expression in terms
of the specified variables.
EXECUTE:
(a)
E
is related to
V
and
J
is related to
I,
so use
P = VI.
This gives
VI
p
LA
=
and
so
VI
E
JpE
J
LA
=
(b)
J
is related to
I
and
ρ
is related to
R,
so use
2
.
PI
R
=
This gives
2
.
IR
p
LA
=
22
2
2
and
so
LJ
A
L
IJ
A R
p
J
AL
A
ρρ
=
(c)
E
is related to
V
and
is related to
R
, so use
2
/.
PV R
=
This gives
2
.
V
p
RLA
=
2
and
so
.
L
EL
A
E
VE
L R
p
A
L
=
=
EVALUATE:
For a given material (
constant),
p
is proportional to
or to
.
J
E
25.48.
IDENTIFY:
Calculate the current in the circuit.
The power output of a battery is its terminal voltage times the
current through it.
The power dissipated in a resistor is
2
.
SET UP:
The sum of the potential changes around the circuit is zero.
EXECUTE:
(a)
8.0 V
0.47 A
17
I
Ω
.
Then
9
(0.47 A) (5.0
) 1.1 W
R
Ω
Ω
=
and
9
(0.47 A) (9.0
)
2.0 W.
R
Ω
Ω
=
(b)
16V
(16 V)(0.47 A)
(0.47 A) (1.6
)
7.2 W.
I
r
=− =
−
Ω
=
E
(c)
8V
(8.0 V)(0.47 A)
(0.47 A) (1.4 )
4.1W.
I
r
=+ =
+
Ω
=
E
EVALUATE:
(d)
(b)
(a) (c)
=+
.
The rate at which the 16.0 V battery delivers electrical energy to the circuit
equals the rate at which it is consumed in the 8.0 V battery and the 5.0
Ω
and 9.0
Ω
resistors.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentCurrent, Resistance, and Electromotive Force
2511
25.49.
(a) IDENTIFY
and
SET UP:
PV
I
=
and energy = (power)
×
(time).
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Current, Resistance, Conservation Of Energy, Energy, Volt, electrical energy

Click to edit the document details