HW-9Solutions-02-08-08

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

• Notes
• 5

This preview shows pages 1–2. Sign up to view the full content.

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 V iR = = (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 V iR = = (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?

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern