756_PartUniversity Physics Solution

# 756_PartUniversity Physics Solution - 25-20 Chapter 25 The...

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25-20 Chapter 25 SET UP: The circuit is sketched in Figure 25.79. EXECUTE: ( ) 12 0 Ir r R −+ + = EE I rrR = + + 12.0 V 8.0 V 1.0 1.0 8.0 I = Ω 0.40 A I = Figure 25.79 (b) () ( ) ( ) 2 222 2 1 2 0.40 A 8.0 1.0 1.0 PIRI rI r IRrr = ++ = + += Ω 1.6 W P = (c) Chemical energy is converted to electrical energy in a battery when the current goes through the battery from the negative to the positive terminal, so the electrical energy of the charges increases as the current passes through. This happens in the 12.0 V battery, and the rate of production of electrical energy is 1 12.0 V 0.40 A 4.8 W. PI == = E (d) Electrical energy is converted to chemical energy in a battery when the current goes through the battery from the positive to the negative terminal, so the electrical energy of the charges decreases as the current passes through. This happens in the 8.0 V battery, and the rate of consumption of electrical energy is ( ) 2 8.0 V 0.40 V 3.2 W. = E (e) EVALUATE: Total rate of production of electrical energy = 4.8 W. Total rate of consumption of electrical energy = 1.6 W + 3.2 W = 4.8 W, which equals the rate of production, as it must. 25.80. IDENTIFY: Apply L R A ρ = for each material. The total resistance is the sum of the resistances of the rod and the wire. The rate at which energy is dissipated is 2 . IR SET UP: For steel, 7 2.0 10 m =× Ω . For copper, 8 1.72 10 m. = ×Ω EXECUTE: (a) 7 3 steel 2 (2.0 10 m) (2.0 m) 1.57 10 ( 4) (0.018 m) L R A π = × Ω and 8 Cu 2 (1.72 10 m) (35 m) 0.012 ( 4) (0.008 m) L R A = Ω . This gives 3 steel Cu ( ) (15000 A) (1.57 10 0.012 ) 204 V. VI RIR R + = × Ω + Ω = (b) 22 6 (15000 A) (0.0136 ) (65 10 s) 199 J. EP tIR t == = Ω × = EVALUATE: 2 is large but t is very small, so the energy deposited is small. The wire and rod each have a mass of about 1 kg, so their temperature rise due to the deposited energy will be small. 25.81. IDENTIFY and SET UP: The terminal voltage is ab r I R ε = −= , where R is the resistance connected to the battery. During the charging the terminal voltage is ab r = + . PV I = and energy is E Pt = . 2 Ir is the rate at which energy is dissipated in the internal resistance of the battery. EXECUTE: (a) 12.0 V (10.0 A) (0.24 ) 14.4 V. ab r =+ = + Ω= (b) 6 (10 A) (14.4 V) (5) (3600 s) 2.59 10 J. tI V t = × (c) 5 diss diss (10 A) (0.24 ) (5) (3600 s) 4.32 10 J. t I r t === Ω (d) Discharged at 10 A: 12.0 V (10 A) (0.24 ) 0.96 . 10 A Ir rR I −− Ω =⇒ = = = Ω + E E (e) 6 (10A)(9.6V)(5)(3600s) 1.73 10 J. V t (f) Since the current through the internal resistance is the same as before, there is the same energy dissipated as in (c): 5 diss 4.32 10 J. E (g) Part of the energy originally supplied was stored in the battery and part was lost in the internal resistance. So the stored energy was less than what was supplied during charging. Then when discharging, even more energy is lost in the internal resistance, and only what is left is dissipated by the external resistor.

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Current, Resistance, and Electromotive Force 25-21 25.82. IDENTIFY and SET UP: The terminal voltage is , ab VI r I R ε =−= where R is the resistance connected to the battery. During the charging the terminal voltage is .
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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756_PartUniversity Physics Solution - 25-20 Chapter 25 The...

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