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Unformatted text preview: Problem 5. A 1.5V battery stores 4.5 kJ of energy. How long can it light a flashlight bulb that draws 0.60 A? Solution The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For an ideal battery, P = E I , therefore E It = 4.5 kJ, or t = 4.5 kJ = (1.5 V)(0.60 A) = 5 × 10 3 s = 1.39 h. Problem 10. In Fig. 2849 all resistors have the same value, R . What will be the resistance measured (a) between A and B or (b) between A and C ? FIGURE 2849 Problems 10 and 11. Solution (a) The resistance between A and B is equivalent to two resistors of value R in series with the parallel combination of resistors of values R and 2 R . Thus, R AB = R + R + R (2 R ) = ( R + 2 R ) = 8 R = 3. (b) R AC is equivalent to just one resistor of value R in series with the parallel combination of R and 2 R (since the resistor at point B carries no current, i.e., its branch is an open circuit). Thus R AC = R + R (2 R ) = 3 R = 5 R = 3. Problem 12. A defective starter motor in a car draws 300 A from the car’s 12V battery, dropping the battery terminal voltage to only 6 V. A good starter motor should draw only 100 A. What will the battery terminal voltage be with a good starter? Solution The starter circuit contains all the resistances in series, as in Fig. 2810. (We assume R L includes the resistance of the cables, connections, etc., as well as that of the motor.) With the defective starter, V T = E − IR int = 6 V = 12 V − (300 A) R int , so R int = 0.02 Ω . With a good starter, V T = 12 V − (100 A)(0.02 Ω ) = 10 V. Problem 19. What is the equivalent resistance between A and B in each of the circuits shown in Fig. 2850? Hint: In ( c ), think about symmetry and the current that would flow through R 2 . Solution (a) There are two parallel pairs ( 1 2 R 1 ) in series, so R AB = 1 2 R 1 + 1 2 R 1 = R 1 . (b) Here, there are two series pairs (2 R 1 ) in parallel, so R AB = ( 2 R 1 )( 2 R 1 ) = ( 2 R 1 + 2 R 1 ) = R 1 . (c) Symmetry requires that the current divides equally on the right and left sides, so points C and D are at the same potential. Thus, no current flows through R 2 , and the circuit is equivalent to (b). (Note that the reasoning in parts (a) and (b) is easily generalized to resistances of different values; the generalization in part (c) requires the equality of ratios of resistances which are mirror images in the plane of symmetry.) CHAPTER 28 665 FIGURE 2850 Problem 19 Solution....
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This note was uploaded on 02/16/2010 for the course PHYS 2B 2b taught by Professor Hirsch during the Winter '10 term at UCSD.
 Winter '10
 Hirsch

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