spring_08_exam2_sol_phy2054

spring_08_exam2_sol_phy2054 - PHY2054 Spring 2008 Prof. P....

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PHY2054 Spring 2008 1 P r o f . P . K u m a r P r o f . P . A v e r y March 5, 2008 Exam 2 Solutions 1. Two cylindrical resistors are made of the same material and have the same resistance. The resistors, R 1 and R 2 , have different radii, r 1 and r 2 , and different lengths, L 1 and L 2 . Which of the following relative values for radii and lengths would result in equal resistances? (1) 2 r 1 = r 2 and 4 L 1 = L 2 (2) 2 r 1 = r 2 and L 1 = 2 L 2 (3) r 1 = r 2 and 4 L 1 = L 2 (4) r 1 = r 2 and L 1 = 2 L 2 (5) none of these The resistance is given by / R LA ρ = , where is the resistivity, L is the length and A is the cross sectional area. To keep R unchanged, the ratio of L to A must be unchanged. Since A = π r 2 , if r is doubled then L must be multiplied by 4. Only the answer (1) works. 2. If a 9.0-V battery with internal resistance 0.2 Ω and an 18 Ω resistor are connected together, what is the amount of electrical energy (in J) transformed to heat per coulomb of charge that flows through the circuit? (1) 9.0 (2) 3.0 (3) 0.50 (4) 72 (5) 4.5 The amount of heat released is P Δ t, where P = Vi. The total amount of charge is q = i t. Thus the amount of heat energy released per coulomb of charge is P t / i t = P/i = V = 9V.
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PHY2054 Spring 2008 2 3. ( 2054 exam, #1, 2006 ) At room temperature conditions of 25 ° C, a circuit containing a metal resistor connected to a 50 V battery has a current of 2.00 A. When the wire is heated to 145 ° C the current drops to 1.94 A. What is the temperature coefficient of resistivity of this metal? (1) 0.00026 / ° C (2) 0.0064 / ° C (3) 0.031 / ° C (4) 0.00075 / ° C (5) 0.077 / ° C The resistance at 25 ° C is 50 / 2.00 = 25 Ω . The resistance at 145 ° C is 50 / 1.94 = 25.773 . Thus the temperature coefficient of resistivity is 0.773 / 25.0 / 120 ° = 0.00026 / ° C. 4. ( Chap. 18.9 ) Consider the circuit shown here, with R = 15 Ω . What is the potential difference between points a and b ? (1) 5.56 V (2) 19.4 V (3) 25 V (4) 12.3 V (5) None of these This circuit is equivalent to a 10 resistor in series with a parallel combination of 10 , 5 and 20 , yield a total resistance of 12.86 . Thus the total current is 25 / 12.86 = 1.944 A. The voltage difference is thus 25 – 1.944*10 = 5.56 V. 5. ( 2049 Exam 2, #4, Fall ’07 ) Consider the circuit shown in the accompanying figure. What is the power dissipated (in watts) in the 5 resistor? (1) 0.43 (2) 0.11 (3) 0.28 (4) 2.35 (5) 0.069 The two equations from Kirchoff’s rules are ( ) 11 2 12 5 0 ii i −+ = and () 21 2 25 0 i = .
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This test prep was uploaded on 04/17/2008 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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spring_08_exam2_sol_phy2054 - PHY2054 Spring 2008 Prof. P....

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