Physics Solution Manual for 1100 and 2101

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Unformatted text preview: and b to denote the inner and outer radii, respectively, and using Equation 20.3 to express the resistance for each portion, we find for the equivalent resistance that ( 2 2 ACu AAl 1 1 1 π a2 π b − a = + = + = + Rp RAl RCu ρCu L ρ Al L ρ Cu L ρ Al L = ( π 2.00 ×10−3 m (1.72 ×10 −8 ) ) 2 Ω ⋅ m (1.50 m ) + ( ) )( ) π 3.00 ×10−3 m − 2.00 ×10−3 m 2 ( 2.82 ×10−8 Ω ⋅ m ) (1.50 m ) 2 = 0.00116 Ω We have taken resistivity values for copper and aluminum from Table 20.1 ______________________________________________________________________________ Chapter 20 Problems 1091 61. REASONING Since the defogger wires are connected in parallel, the total resistance of all thirteen wires can be obtained from Equation 20.17: 1 13 = Rp R Rp = or R 13 where R is the individual resistance of one of the wires. The heat required to melt the ice is given by Q = mLf , where m is the mass of the ice and Lf is the latent heat of fusion of the ice (see Section 12.8). Therefore, using Equation 20.6c, we can see that the power or energy dissipated per unit time in the wires and used to melt the ice is P= V2 mLf = Rp 1 24 4t 3 or V2 mLf = R /13 t Energy/time According to Equation 20.3, R = ρ L / A , where the length of the wire is L, its crosssectional area is A and its resistivity is ρ; therefore, the last expression can be written mLf 13V 2 13V 2 = = R ρL / A t This expression can be solved for the area A. SOLUTION Solving the above expression for A, and substituting given data, and obtaining the latent heat of fusion of water from Table 12.3, we find that A= = ρ LmLf 13V 2t (88.0 ×10 –8 Ω ⋅ m)(1.30 m)(2.10 × 10 –2 kg)(33.5 × 104 J/kg) 2 = 3.58 × 10 –8 m 2 13(12.0 V) (120 s) ______________________________________________________________________________ 62. REASONING Between points a and b there is only one resistor, so the equivalent resistance is Rab = R. Between points b and c the two resistors are in parallel. The equivalent resistance can be found from Equation 20.17: 1 112 =+= Rbc R R R so Rbc = 1 R 2 The equivalent resistance between a and b is in series with the equivalent resistance between b and c, so the equivalent resistance between a and c is 1092 ELECTRIC CIRCUITS 3 Rac = Rab + Rbc = R + 1 R = 2 R 2 Thus, we see that Rac > Rab > Rbc. SOLUTION Since the resistance is R = 10.0 Ω, the equivalent resistances are: Rab = R = 10.0 Ω Rbc = 1 R = 5.00 Ω 2 Rac = 3 R = 15.0 Ω 2 ______________________________________________________________________________ 63. SSM 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 –1 –1 combination can be obtained from Equation 20.17 as R p = ( R coffee + Rpan ) –1 + R bread . –1 SOLUTION Using Ohm’s law and the expression developed above for Rp , we find I= V 1 1 1 1 = (120 V ) = V + + = 9.2 A R Rp 14 Ω + 16 Ω 23 Ω coffee + R pan R bread ______________________________________________________________________________ 64. REASONING We will approach this problem in parts. The resistors that are in series will be combined according to Equation 20.16, and the resistors that are in parallel will be combined according to Equation 20.17. SOLUTION The 1.00 Ω, 2.00 Ω and 3.00 Ω resistors are in series with an equivalent resistance of Rs = 6.00 Ω. 2.00 Ω 4.00 Ω 6.00 Ω 3.00 Ω 6.00 Ω Chapter 20 Problems This equivalent resistor of 6.00 is in parallel with the 3.00-Ω resistor, so 1 1 1 = + RP 6.00 Ω 3.00 Ω 2.00 Ω 4.00 Ω 1093 6.00 Ω 2.00 Ω RP = 2.00 Ω 2.00 Ω This new equivalent resistor of 2.00 is in series with the 6.00Ω resistor, so Rs' = 8.00 Ω. Rs' is in parallel with the 4.00-Ω resistor, so 1 RP′ = 1 1 + 8.00 Ω 4.00 Ω 4.00 Ω 8.00 Ω 2.00 Ω 2.67 Ω RP′ = 2.67 Ω Finally, Rp' is in series with the 2.00-Ω, so the total equivalent resistance is 4.67 . ______________________________________________________________________________ 65. SSM REASONING When two or more resistors are in series, the equivalent resistance is given by Equation 20.16: Rs = R1 + R2 + R3 + . . . . Likewise, when resistors are in parallel, the expression to be solved to find the equivalent resistance is given by Equation 1 1 1 1 20.17: = + + + ... . We will successively apply these to the individual resistors Rp R1 R2 R3 in the figure in the text beginning with the resistors on the right side of the figure. SOLUTION Since the 4.0-Ω and the 6.0-Ω resistors are in series, the equivalent resistance of the combination of those two resistors is 10.0 Ω. The 9.0-Ω and 8.0-Ω resistors are in parallel; their equivalent resistance is 4.24 Ω. The equivalent resistances of the parallel combination (9.0 Ω and 8.0 Ω...
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