Download Assignment 3

CHAPTER 20 ELECTRIC CIRCUITS
PROBLEMS
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1.
SSM REASONING Since current is defined as charge per unit time, the current used by the
portable compact disc player is equal to the charge provided by the battery pack (180 C) divided by
the time in which the charge is delivered (2.0 h).
SOLUTION The amount of current that the player uses in operation is determined from Equation
20.1:
I=
∆q 180 C  1.0 h 
=
∆t
3600
s
2.0 h 1
4
24
3
= 0.025 A
Converts hours
to seconds
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8.
REASONING As discussed in Section 20.1, the voltage gives the energy per unit charge. Thus, we
can determine the energy delivered to the toaster by multiplying the voltage V by the charge ∆q that
flows during a time ∆t of one minute. The charge can be obtained by solving Equation 20.1,
I = (∆q)/(∆t), since the current I can be obtained from Ohm’s law.
SOLUTION Remembering that voltage is energy per unit charge, we have
Energy = V ∆q
Solving Equation 20.1 for ∆q gives ∆q = I ∆t, which can be substituted in the previous result to give
Energy = V ∆q = VI ∆t
According to Ohm’s law (Equation 20.2), the current is I = V/R, which can be substituted in the
energy expression to show that
2
V 2 ∆t (120 V ) ( 60 s )
V
Energy = VI ∆t = V   ∆t =
=
= 6.2 × 10 4 J
 R
14 Ω
R
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15. REASONING AND SOLUTION The resistance of the cable is
R=
V ρL
=
I
A
Since A = π r2, the radius of the cable is
r=
ρ LI
=
πV
(1.72 × 10 –8 Ω ⋅ m )(0.24 m )(1200 A ) =
π (1.6 × 10–2 V )
9.9 × 10 –3 m
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Chapter 20 Problems 607
19. REASONING AND SOLUTION Suppose that when the initial temperature of the wire is T 0 the
resistance is R 0, and when the temperature rises to T the resistance is R. The relation between
temperature and resistance is given by Equation 20.5 as R = R0[1 + α (T – T0 )], where α is the
temperature coefficient of resistivity. The initial and final resistances are related to the voltage and
current as R0 = V/I0 and R = V/I, where the voltage V across the wire is the same in both cases.
Substituting these values for R0 and R into Equation 20.5 and solving for T, we arrive at
T = T0 +
 I0

− 1

 I

α
= 20 ° C +
 1.50 A 
− 1

 1.30 A 
4.5 × 10 −4 (C° ) −1
= 360 ° C
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23. REASONING AND SOLUTION According to Equation 20.6c, the power dissipated by the iron is
2
V 2 (120 V )
P=
=
= 6.0 × 10 2 W
R
24 Ω
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25. SSM REASONING According to Equation 6.10, 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 drier and then to the
computer.
SOLUTION The energy used by the drier is
Energy = Pt = IVt = (16 A)(240 V)(45 min)
 60 s 
7
= 1.04 × 10 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 × 10 7 J
 1.00 h 
t=
= 3.21 × 10 4 s = 3.21 × 10 4 s 
= 8.9 h
 3600 s 
(2.7 A )(120 V )
(
)
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27. REASONING AND SOLUTION We know P = V2/R.
a.
V=
PR = (0.25 W)(680 Ω) = 13 V
V=
PR = (2.0 W)(680 Ω) = 37 V
b.
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608
ELECTRIC CIRCUITS
2 /R
34. REASONING The average power is given by Equation 20.15c as P = Vrms
. In this expression
the rms voltage Vrms appears. However, we seek the peak voltage V0. The relation between the two
types of voltage is given by Equation 20.13 as Vrms = V0 / 2 , so we can obtain the peak voltage by
using Equation 20.13 to substitute into Equation 20.15c.
SOLUTION Substituting Vrms from Equation 20.13 into Equation 20.15c gives
P=
2
Vrms
R
(V0 /
=
2)
2
R
=
V02
2R
Solving for the peak voltage V0 gives
V0 =
2 RP =
2 ( 4.0Ω )( 55 W ) = 21 V
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42. REASONING AND SOLUTION The equivalent resistance of the circuit is
Rs = R1 + R2 = 36.0 Ω + 18.0 Ω = 54.0 Ω
Ohm's law for the circuit gives
I = V/Rs = (15.0 V)/(54.0 Ω) = 0.278 A
a. Ohm's law for R1 gives
V1 = (0.278 A)(36.0 Ω) = 10.0 V
b. Ohm's law for R2 gives
V2 = (0.278 A)(18.0 Ω) = 5.00 V
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44. REASONING AND SOLUTION
a. The equivalent resistance of the circuit is
Rs = 9.0 Ω + 5.0 Ω + 1.0 Ω = 15.0 Ω
The current through each of the resistors is from Ohm's law
I = (24 V)/(15.0 Ω) = 1.6 A
b. The voltage drop across the 9.0-Ω resistor is
V1 = (1.6 A)(9.0 Ω) = 14 V
Chapter 20 Problems 609
The drop across the 5.0 Ω resistor is
V2 = (1.6 A)(5.0 Ω) = 8.0 V
The drop across the 1.0 Ω resistor is
V3 = (1.6 A)(1.0 Ω) = 1.6 V
c. The power dissipated in the 9.0 Ω resistor is
P1 = I2 R1 = (1.6 A)2(9.0 Ω) = 23 W
Similarly, for the 5.0 Ω resistor
P2 = (1.6 A)2(5.0 Ω) = 13 W
and for the 1.0 Ω resistor
P3 = (1.6 A)2(1.0 Ω) = 2.6 W
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50. REASONING AND SOLUTION The rule for combining parallel resistors is
1
1
1
=
+
R P R1 R 2
which gives
1
1
1
1
1
=
−
=
−
or
R 2 = 446 Ω
R P R1 115 Ω 155 Ω
R2
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53. SSM REASONING Since the resistors are connected in parallel, the voltage across each one is
the same and can be calculated from Ohm's Law (Equation 20.2: V = IR ). Once the voltage across
each resistor is known, Ohm's law can again be used to find the current in the second resistor. The
total power consumed by the parallel combination can be found calculating the power consumed by
each resistor from Equation 20.6b: P = I2R. Then, the total power consumed is the sum of the power
consumed by each resistor.
SOLUTION Using data for the second resistor, the voltage across the resistors is equal to
V = IR = (3.00 A)(64.0 Ω) = 192 Ω
a. The current through the 42.0-Ω resistor is
I=
V 192 V
=
= 4.57 A
R 42.0 Ω
b. The power consumed by the 42.0-Ω resistor is
610
ELECTRIC CIRCUITS
P = I 2 R = ( 4.57 A) 2 (42.0 Ω) = 877 W
while the power consumed by the 64.0-Ω resistor is
P = I 2 R = (3.00 A) 2 (64.0 Ω ) = 576 W
Therefore the total power consumed by the two resistors is 877 W + 576 W = 1450 W
.
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57. SSM REASONING The equivalent resistance of the three devices in parallel is Rp , and we can
find the value of Rp by using our knowledge of the total power consumption of the circuit; the value
of Rp can be found from Equation 20.6c, P = V 2 / Rp . Ohm's law (Equation 20.2, V = IR ) can then
be used to find the current through the circuit.
SOLUTION
a. The total power used by the circuit is P = 1650 W + 1090 W +1250 W = 3990 W. The equivalent
resistance of the circuit is
V 2 (120 V)2
Rp =
=
= 3.6 Ω
P
3990 W
b. The total current through the circuit is
I=
V
120 V
=
= 33 A
R p 3.6 Ω
The total current is larger than the rating of the circuit breaker; therefore, the breaker will open .
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60. REASONING To find the current, we will use Ohm’s law, together with the proper equivalent
resistance. The coffee maker and frying pan are in series, so their equivalent resistance is given by
Equation 20.16 as Rcoffee + Rpan. This total resistance is in parallel with the resistance of the bread
maker, so the equivalent resistance of the parallel combination can be obtained from Equation 20.17
–1
as R p–1 = (R coffee + Rpan ) –1 + R bread
.
SOLUTION Using Ohm’s law and the expression developed above for Rp–1, we find
I=


V
1
1 
1 
1
 = (120 V )
= V 
+
+
 = 9.2 A
Rp
 14 Ω + 16 Ω 23 Ω 
 R coffee + R pan R bread 
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67. REASONING AND SOLUTION The resistors in the small network have an equivalent resistance
Chapter 20 Problems 611
1
1
1
=
+
R p 2.0 Ω + 1.0 Ω 5.0 Ω + 1.0 Ω
or
R p = 2.0 Ω
This resistance is in series with the 4.0-Ω resistor so the equivalent resistance of the circuit is
R = 6.0 Ω. Therefore, Ohm's law gives the total current in the circuit to be
I = V/R = (12 V)/(6.0 Ω) = 2.0 A
This current, upon entering the parallel branch, will split in the ratios of 3:9 and 6:9, with the largest
current entering the smallest resistance path. The 5.0-Ω resistor then has a current of
I = (3/9)(2.0 A)
The power dissipated in this resistor is
P = [(3/9)(2.0 A)]2(5.0 Ω) = 2.2 W
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68. REASONING AND SOLUTION The equivalent resistance of the initial configuration is given by
1/Rp = 3/R
or
Rp = R/3
The parallel part of the final configuration has a resistance of
Rp' = R/2
so the total equivalent resistance is
Rs = R/2 + R = 3R/2
Now
so
Rs = Rp + 700 Ω
3R/2 = R/3 + 700 Ω
which gives
R = 600 Ω
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