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Unformatted text preview: V 12 V
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I= 24. REASONING AND SOLUTION The power delivered is P = VI, so that we have
a. Pbd = VIbd = (120 V)(11 A) = 1300 W b. Pvc = VIvc = (120 V)(4.0 A) = 480 W c. The energy is E = Pt, so that we have Ebd Pbd tbd (1300 W)(15 min)
= 1.4
Evc Pvctvc (480 W)(30.0 min)
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= = 25. REASONING The total cost of keeping all the TVs turned on is equal to the number of
TVs times the cost to keep each one on. The cost for one TV is equal to the energy it
consumes times the cost per unit of energy ($0.12 per kW⋅h). The energy that a single set
uses is, according to Equation 6.10b, the power it consumes times the time of use.
SOLUTION The total cost is
Total cost = (110 million sets ) ( Cost per set ) $0.12 = (110 million sets ) Energy ( in kW ⋅ h ) used per set 1 kW ⋅ h The energy (in kW⋅h) used per set is the product of the power and the time, where the power
is expressed in kilowatts and the time is in hours: 1 kW Energy used per set = P t = ( 75 W ) ( 6.0 h ) 1000 W (6.10b) The total cost of operating the TV sets is $0.12 1 kW 6
Total cost = (110 million sets ) ( 75 W ) ( 6.0 h ) = $5.9 × 10
1000 W 1 kW ⋅ h ______________________________________________________________________________ 1068 ELECTRIC CIRCUITS 26. REASONING
2
a. The power delivered to a resistor is given by Equation 20.6c as P = V / R , where V is
2 the voltage and R is the resistance. Because of the dependence of the power on V ,
doubling the voltage has a greater effect in increasing the power than halving the
resistance. The following table shows the power for each circuit, given in terms of these
variables, and confirms this fact. The table also gives the expected ranking, in decreasing
order, of the power.
Power
a b
c d V2
P=
R
2
V
P=
2R
2
( 2V )
P=
=
R
2
( 2V )
P=
=
2R Rank
3
4 4V 2
R 1 2V 2
R 2 b. The current is given by Equation 20.2 as I = V/R. Note that the current, unlike the
power, depends linearly on the voltage. Therefore, either doubling the voltage or halving
the resistance has the same effect on the current. The following table shows the current for
the four circuits and confirms this fact. The table also gives the expected ranking, in
decreasing order, of the current. Current Rank a I= V
R 2 b I= V
2R 3 c I= 2V
R 1 2V V
=
2R R 2 d I= Chapter 20 Problems 1069 SOLUTION
a. Using the results from the REASONING and the values of V = 12.0 V and R = 6.00 Ω,
we find that the power dissipated in each resistor is
Power Rank a 2
(12.0 V )
V
P=
=
= 24.0 W
R
6.00 Ω 3 b (12.0 V )
V2
P=
=
= 12.0 W
2 R 2 ( 6.00 Ω ) 4 c 4V 2 4 (12.0 V )
P=
=
= 96.0 W
R
( 6.00 Ω ) 1 d 2
2 (12.0 V )
2V
P=
=
= 48.0 W
R
( 6.00 Ω ) 2 2 2 2 2 b. Using the results from part (b) and the values of V = 12.0 V and R = 6.00 Ω, we find
that the current in each circuit is
Current
a
b V 12.0 V
=
= 2.00 A
R 6.00 Ω
12.0 V
V
I=
=
= 1.00 A
2 R 2 ( 6.00 Ω )
I= Rank
2 3 2V 2 (12.0 V )
=
= 4.00 A
1
R
6.00 Ω
2V 2 (12.0 V )
I=
=
= 2.00 A
d
2
2 R 2 ( 6.00 Ω )
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c 27. I= SSM REASONING According to Equation 6.10b, the energy used is Energy = Pt, where
P is the power and t is the time. According to Equation 20.6a, the power is P = IV, where I
is the current and V is the voltage. Thus, Energy = IVt, and we apply this result first to the
dryer and then to the computer.
SOLUTION The energy used by the dryer is 1070 ELECTRIC CIRCUITS Energy = Pt = IVt = (16 A)(240 V)(45 min) 60 s = 1.04 × 107 J 1.00 min 14243
Converts minutes
to seconds For the computer, we have Energy = 1.04 × 10 7 J = IVt = (2.7 A )(120 V)t
Solving for t we find ( ) 1.04 × 107 J 1.00 h = 3.21 × 10 4 s = 3.21 × 104 s = 8.9 h
( 2.7 A )(120 V ) 3600 s ______________________________________________________________________________
t= 28. REASONING AND SOLUTION We know that the resistance of the wire can be obtained
from
2
2
P = V /R
or
R = V /P
We also know that R = ρL/A. Solving for the length, noting that A = π r , and using
2 ρ = 100 × 10–8 Ω.m from Table 20.1, we find
L= RA ρ (V 2 / P ) (π r 2 ) = V 2π r 2 =
=
ρ ρP (120 V )2 π ( 6.5 × 10 –4 m ) 2 (100 × 10–8 Ω ⋅ m )( 4.00 × 102 W ) = 50 m ______________________________________________________________________________
29. REASONING A certain amount of time t is needed for the heater to deliver the heat Q
required to raise the temperature of the water, and this time depends on the power produced
by the heater. The power P is the energy (heat in this case) per unit time, so the time is the
heat divided by the power or t = Q/P. The heat required to raise the temperature of a mass m
of water by an amount ∆T is given by Equation 12.4 as Q = cm∆T, where c is the specific
heat capacity of water [4186 J/(kg·Cº), see Table 12.2]. The power dissipated in a resistance
R is given by Equation 20.6c as P = V 2 / R , where V is the voltage across the resistor. Using...
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 Spring '13
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 Physics, The Lottery

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