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Physics Solution Manual for 1100 and 2101

# 00 a640 192 v a the current through the 420 resistor

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Unformatted text preview: 2 Vrms R (Equation 20.15c). SOLUTION When plugged into an outlet in the US that has an rms voltage Vrms = 120 V, the average power output P of the heater is, from Equation 20.15c, P= 2 Vrms R (20.15c) When operated in Germany, the average power output ( P0 = 550 W ) of the heater is given by P0 = obtain 2 Vrms,0 R (Equation 20.15c), where Vrms,0 = 230 V. Solving this relation for R, we Chapter 20 Problems R= 2 Vrms,0 P0 1075 (1) Substituting Equation (1) into Equation 20.15c yields P= 2 Vrms R (120 V )2 550 W = 150 W = = 2 P= ( ) 2 Vrms,0 Vrms,0 0 ( 230 V )2 2 Vrms 2 Vrms P0 ______________________________________________________________________________ 37. REASONING AND SOLUTION The power P dissipated in the extension cord is P = I 2 R (Equation 20.6b). The resistance R is related to the length L of the wire and its crosssectional area A by Equation 20.3, R = ρL/A, where ρ is the resistivity of copper. The crosssectional area of the wire can be expressed as A= ρL R = ρI2 P L where the ratio P/L is the power per unit length of copper wire that the heater produces. The wire is cylindrical, so its cross-sectional area is A = π r2 . Thus, the smallest radius of wire that can be used is r=I ρ P L π = (18 A) 1.72 × 10−8 Ω ⋅ m = 1.3 × 10−3 m π (1.0 W / m) Note that we have used 1.0 W/m as the power per unit length, rather than 2.0 W/m. This is because an extension cord is composed of two copper wires. If the maximum power per unit length that the extension cord itself can produce is 2.0 W/m, then each wire can produce only a maximum of 1.0 W/m. ______________________________________________________________________________ 38. REASONING The higher the average power output P of the resistance heater, the faster it Q can supply the heat Q needed to raise the water’s temperature. This follows from P = t Q (Equation 6.10b), which implies that the recovery time is t = . The average power output P 2 V P of the resistance heater is given by P = rms (Equation 20.15c), where R is the resistance R 1076 ELECTRIC CIRCUITS of the heater and Vrms = 120 V is the heater’s rms operating voltage. The heater must supply an amount of heat Q = cm∆T (Equation 12.4) to cause an increase ∆T in the temperature of a mass m of water, where c = 4186 J/ ( kg ⋅ Co ) is the specific heat capacity of water (see Table 12.2). The mass m of the water can be found from the density ρ = 1.000×103 kg/m3 (see Table 11.1) and the volume Vwater of the water held in the heater, according to m = ρVwater (Equation 11.1). SOLUTION Substituting P = the expression t = 2 Vrms R (Equation 20.15c) and Q = cm∆T (Equation 12.4) into Q for the recovery time, we obtain P t= Q cm∆T Rcm∆T = = 2 P V 2 Vrms rms R (1) Substituting m = ρVwater (Equation 11.1) into Equation (1) yields t= Rcm∆T 2 Vrms = Rc ρVwater ∆T 2 Vrms (2) Before using Equation (2) to calculate the recovery time, we must convert the volume Vwater of the water in the unit from gallons to m3, using the equivalence 1 gal = 3.79×10−3 m3: 3.79 ×10−3 m3 3 Vwater = ( 52 gal ) = 0.20 m 1.00 gal We note that the recovery time is to be expressed in hours, where 1.00 h = 3600 s. Therefore, from Equation (2), we find that t= Rc ρVwater ∆T 2 Vrms = ( 3.0 Ω ) 4186 J/ ( kg ⋅ Co ) (1.000 ×103 kg/m3 ) ( 0.20 m3 ) ( 53 oC − 11 oC ) (120 V )2 1.00 h = 7300 s = ( 7300 s ) = 2.0 h 3600 s ______________________________________________________________________________ Chapter 20 Problems 39. 1077 SSM REASONING a. We can obtain the frequency of the alternating current by comparing this specific expression for the current with the more general one in Equation 20.8. b. The resistance of the light bulb is, according to Equation 20.14, equal to the rms-voltage divided by the rms-current. The rms-voltage is given, and we can obtain the rms-current by dividing the peak current by 2 , as expressed by Equation 20.12. c. The average power is given by Equation 20.15a as the product of the rms-current and the rms-voltage. SOLUTION a. By comparing I = ( 0.707 A ) sin ( 314 Hz ) t with the general expression (see Equation 20.8) for the current in an ac circuit, I = I 0 sin 2 π f t , we see that 2π f t = ( 314 Hz ) t or 314 Hz = 50.0 Hz 2π f= b. The resistance is equal to Vrms/Irms, where the rms-current is related to the peak current I0 by I rms = I 0 / 2 . Thus, the resistance of the light bulb is R= Vrms I rms = Vrms 2 (120.0 V ) = = 2.40 × 102 Ω I0 0.707 A 2 (20.14) c. The average power is the product of the rms-current and rms-voltage: I 0.707 A P = I rmsVrms = 0 Vrms = (20.15a) (120.0 V ) = 60.0 W 2 2 ______________________________________________________________________________ 40. REASONING AND SOLUTION The energy Q1 that is released when the water cools from an initial temperature T to a final temperature of 0.0 °C is given by Equation 12.4 as Q1 = cm(T – 0.0 °C). The energy Q2 released when the water turns into ice at 0.0 °C is Q2 = mLf, where Lf is the latent heat of fusion for water. Since power P is energy divided by time, the power produced is P= Q1 + Q...
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