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Unformatted text preview: Physics 204, morning summer 2008 Homework 5 solutions Chapter 20 Problems 17,45,55,64 17. REASONING AND SOLUTION Suppose that when the initial temperature of the wire is T0 the resistance is R0, 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 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 45. SSM REASONING The resistance of one of the wires in the extension cord is given by Equation 20.3: R = ρ L / A , where the resistivity of copper is ρ = 1.72 × 10−8 Ω.m, according to Table 20.1. Since the two wires in the cord are in series with each other, their total resistance is Rcord = Rwire 1 + Rwire 2 = 2 ρ L / A . Once we find the equivalent resistance of the entire circuit (extension cord + trimmer), Ohm's law can be used to find the voltage applied to the trimmer. SOLUTION a. The resistance of the extension cord is b. Rcord = 2 ρ L 2(1.72 ×10 –8 Ω ⋅ m)(46 m)
=
= 1.2 Ω
A
1.3 × 10 –6 m 2 The total resistance of the circuit (cord + trimmer) is, since the two are in series, Rs = 1.2 Ω + 15.0 Ω = 16.2 Ω Therefore from Ohm's law (Equation 20.2: V = IR), the current in the circuit is I= V 120 V
=
= 7.4 A
Rs 16.2 Ω Finally, the voltage applied to the trimmer alone is (again using Ohm's law), Vtrimmer = (7.4 A)(15.0 Ω) = 110 V 55. 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 2 20.6c, P = V / 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 Rp = b. V 2 (120 V)2
=
= 3.6 Ω
P
3990 W The total current through the circuit is I= V
120 V
=
= 33 A
R p 3.6 Ω This current is larger than the rating of the circuit breaker; therefore, the breaker will open . 64. REASONING The power P dissipated in each resistance R is given by Equation 20.6b as P = I 2 R , where I is the current. This means that we need to determine the current in each resistor in order to calculate the power. The current in R1 is the same as the current in the equivalent resistance for the circuit. Since R2 and R3 are in parallel and equal, the current in R1 splits into two equal parts at the junction A in the circuit. SOLUTION To determine the equivalent resistance of the circuit, we note that the parallel combination of R2 and R3 is in series with R1. The equivalent resistance of the parallel combination can be obtained from Equation 20.17 as follows: 1
1
1
=
+
RP 576 Ω 576 Ω or RP = 288 Ω This 288‐Ω resistance is in series with R1, so that the equivalent resistance of the circuit is given by Equation 20.16 as Req = 576 Ω + 288 Ω = 864 Ω To find the current from the battery we use Ohm’s law: I= 120.0 V
V
=
= 0.139 A 864 Ω
Req Since this is the current in R1, Equation 20.6b gives the power dissipated in R1 as 2
P = I1 R1 = ( 0.139 A ) ( 576 Ω ) = 11.1 W 1
2 R2 and R3 are in parallel and equal, so that the current in R1 splits into two equal parts at the junction A. As a result, there is a current of 1 ( 0.139 A ) in R2 and in R3. Again using Equation 20.6b, we find that the power dissipated in 2
each of these two resistors is 2
⎡2
P2 = I 2 R2 = ⎣ 1 ( 0.139 A ) ⎤
⎦ 2 ( 576 Ω ) = 2.78 W 2
⎡2
⎤
P3 = I 3 R3 = ⎣ 1 ( 0.139 A ) ⎦ 2 ( 576 Ω ) = 2.78 W ...
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 Spring '09
 rollino
 Physics

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