HW_7 Solutions - Physics 2213 25.49. HW #7 Solutions Fall...

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Physics 2213 HW #7 Solutions Fall 2009 25.49. (a) IDENTIFY and SET UP: P VI = and energy = (power) × (time). EXECUTE: ( )( ) 12 V 60 A 720 W P VI = = = The battery can provide this for 1.0 h, so the energy the battery has stored is ( )( ) 6 720 W 3600 s 2.6 10 J U Pt = = = × (b) IDENTIFY and SET UP: For gasoline the heat of combustion is 6 c 46 10 J/kg. L = × Solve for the mass m required to supply the energy calculated in part (a) and use density / mV ρ = to calculate V . EXECUTE: The mass of gasoline that supplies 6 6 6 2.6 10 J is 0.0565 kg. 46 10 J/kg m × ×= = × The volume of this mass of gasoline is 53 33 0.0565 kg 1000 L 6.3 10 m 0.063 L 900 kg/m 1 m m V  = = = ×=   (c) IDENTIFY and SET UP: Energy = (power) × (time); the energy is that calculated in part (a). EXECUTE: 6 , 5800 s 97 min 1.6 h. 450 W U U Pt t P × = = = = = = EVALUATE: The battery discharges at a rate of 720 W (for 60 A) and is charged at a rate of 450 W, so it takes longer to charge than to discharge. 25.81. IDENTIFY and SET UP: The terminal voltage is ab V Ir IR ε =−= , where R is the resistance connected to the battery. During the charging the terminal voltage is ab V Ir = + . P VI = 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 V Ir = + = + Ω= (b) 6 (10 A) (14.4 V) (5) (3600 s) 2.59 10 J. E Pt IVt = = = = × (c) 22 5 diss diss (10 A) (0.24 ) (5) (3600 s) 4.32 10 J. E P t I rt = = = × (d) Discharged at 10 A: 12.0 V (10 A) (0.24 ) 0.96 . 10 A Ir IR rR I −− = ⇒= = = + E E (e) 6 (10 A) (9.6 V) (5) (3600 s) 1.73 10 J. E Pt IVt = = = = × (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. 26.17. IDENTIFY: For resistors in series, the voltages add and the current is the same. For resistors in parallel, the voltages are the same and the currents add. P = I 2 R . (a) SET UP: The circuit is sketched in Figure 26.17a.
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HW_7 Solutions - Physics 2213 25.49. HW #7 Solutions Fall...

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