303L-HW15

303L-HW15 - fierro (jmf2547) – HW15 – berk – (60290)...

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Unformatted text preview: fierro (jmf2547) – HW15 – berk – (60290) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider two cylindrical conductors made out of the same material ( i.e. they have the same density of charge carriers and the same resis- tivity). V 1 vector E 1 I 1 ℓ 1 r 1 b V 2 vector E 2 I 2 ℓ 2 r 2 b If ℓ 2 = 3 ℓ 1 , r 2 = 2 r 1 , V 2 = 4 V 1 , and ρ 2 = ρ 1 , what is the ratio v d, 2 v d, 1 of the magnitudes of the drift velocities? 1. v d, 2 v d, 1 = 1 3 2. v d, 2 v d, 1 = 2 3. v d, 2 v d, 1 = 3 4. v d, 2 v d, 1 = 3 64 5. v d, 2 v d, 1 = 3 4 6. v d, 2 v d, 1 = 64 3 7. v d, 2 v d, 1 = 16 3 8. v d, 2 v d, 1 = 1 2 9. v d, 2 v d, 1 = 3 16 10. v d, 2 v d, 1 = 4 3 correct Explanation: vector J = n qvectorv d Since the conductors are made out of the same material, the density of charge carriers must be the same ( n 1 = n 2 ), so v d, 2 v d, 1 = J 2 n 2 q J 1 n 1 q = J 2 J 1 = ρ V 2 ℓ 1 ρ V 1 ℓ 2 = 4 V 1 ℓ 1 3 V 1 ℓ 1 = 4 3 , since J = E ρ = V l ρ and the two conductors have the same resistivity. 002 (part 2 of 2) 10.0 points What is the ratio R 2 R 1 of the resistances? 1. R 2 R 1 = 3 4 correct 2. R 2 R 1 = 1 3 3. R 2 R 1 = 4 3 4. R 2 R 1 = 1 2 5. R 2 R 1 = 16 3 6. R 2 R 1 = 2 7. R 2 R 1 = 64 3 8. R 2 R 1 = 3 9. R 2 R 1 = 3 16 10. R 2 R 1 = 3 64 Explanation: fierro (jmf2547) – HW15 – berk – (60290) 2 R = ρ ℓ A , so R 2 R 1 = ρ ℓ 2 A 2 ρ ℓ 1 A 1 = ρ ℓ 2 π r 2 2 ρ ℓ 1 π r 2 1 = parenleftbigg r 1 r 2 parenrightbigg 2 ℓ 2 ℓ 1 = parenleftbigg r 1 2 r 1 parenrightbigg 2 3 ℓ 1 ℓ 1 = 3 4 . 003 10.0 points After a 6 . 92 Ω resistor is connected across a battery with a 0 . 26 Ω internal resistance, the electric potential between the physical battery terminals is 7 V....
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303L-HW15 - fierro (jmf2547) – HW15 – berk – (60290)...

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