HW-9Solutions-02-08-08 - Physics 21, Spring 2008 HW-9...

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Physics 21, Spring 2008 February 8, 2008 HW-9 Solutions 9-1. (HRW 27-18) Figure 27-35 shows a resistor of resistance connected to an ideal battery of emf by means of two copper wires. Each wire has length 20.0 cm and radius 1.00 mm. In dealing with such circuits in this chapter, we generally neglect the potential differences along the wires and the transfer of energy to thermal energy in them. Check the validity of this neglect for the circuit of Fig. 27-35 : What is the potential difference across (a) the resistor and (b) each of the two sections of wire? At what rate is energy lost to thermal energy in (c) the resistor and (d) each section of wire? Solution: 18. (a) For each wire, R wire = ρ L/A where A = π r 2 . Consequently, we have R wire = (1.69 × 10 8 m Ω⋅ )(0.200 m)/ π (0.00100 m) 2 = 0.0011 Ω . The total resistive load on the battery is therefore tot R = 2 R wire + R = 2( 0.0011 Ω) + 6.00 Ω = 6.0022 Ω . Dividing this into the battery emf gives the current tot 12.0 V 1.9993 A 6.0022 i R ε == = Ω . The voltage across the R = 6.00 Ω resistor is therefore Vi R (1.9993 A)(6.00 Ω ) = 11.996 V 12.0 V. (b) Similarly, we find the voltage-drop across each wire to be wire wire R (1.9993 A)(0.0011 Ω ) = 2.15 mV. (c) P = i 2 R = (1.9993 A)(6.00 Ω ) 2 = 23.98 W 24.0 W. (d) Similarly, we find the power dissipated in each wire to be 4.30 mW. 9-2. (HRW 27-22) A solar cell generates a potential difference of 0.10 V when a 500 resistor is connected across it, and a potential difference of 0.15 V when a 1000 resistor is substituted. What are the (a) internal resistance and (b) emf of the solar cell?
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This note was uploaded on 02/20/2008 for the course PHYSIC 2 taught by Professor Kim during the Spring '08 term at Lehigh University .

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HW-9Solutions-02-08-08 - Physics 21, Spring 2008 HW-9...

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