HW 15 - yang (ey942) ohw15 turner (56705) 1 This print-out...

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Unformatted text preview: yang (ey942) ohw15 turner (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A charged particle is projected with its initial velocity parallel to a uniform magnetic field. What is the resulting path? 1. circular arc. 2. straight line perpendicular to the field. 3. straight line parallel to the field. correct 4. parabolic arc. 5. spiral. Explanation: The force on a moving charge due to a magnetic field is given by vector F = qvectorv vector B . If vectorv and vector B are parallel, then vectorv vector B = 0 . Hence the force on the particle is zero, and the particle continues to move in a straight line parallel to the field. 002 10.0 points In the diagram below, the resistances are 3 . 3 and 2 . 35 . The current through R 1 is 3 . 74 A. A B R 1 2 R Find the potential difference between points A and B . Correct answer: 21 . 131 V. Explanation: Let : R 1 = 3 . 3 , R 2 = 2 . 35 , and I = 3 . 74 A . Currents through elements in a series combi- nation are the same. The equivalent resis- tance is R = R 1 + R 2 , so V = I R = I ( R 1 + R 2 ) = 3 . 74 A (3 . 3 + 2 . 35 ) = 21 . 131 V . 003 10.0 points The figure represents two possible ways to connect two lighbulbs X and Y to a battery. Bulb X has less resistance than bulb Y . Y X A X Y B Which bulb has the most current running through it? 1. Bulb Y in A 2. Bulb Y in B 3. Bulb X in A correct 4. Bulb X in B Explanation: In A , a parallel circuit, the voltage across X is the same as Y , but X has less resistance, so it has more current running through it. yang (ey942) ohw15 turner (56705) 2 In B , a series circuit, the current is the same through both bulbs. 004 (part 1 of 3) 10.0 points 8 . 3 V 1 . 8 V 4 V I 1 . 4 3 . 3 I 2 5 . 7 I 3 8 . 9 Find the current I 1 in the 0 . 4 resistor at the bottom of the circuit between the two power supplies. Correct answer: 1 . 06788 A. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D At a junction (Conservation of Charge) I 1 + I 2- I 3 = 0 . (1) Kirchhoffs law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoffs law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 0 . 4 , R B = 3 . 3 , R C = 5 . 7 , R D = 8 . 9 , E 1 = 8 . 3 V , E 2 = 1 . 8 V , and E 3 = 4 V . Using determinants, I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1- 1 R A + R B R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Expanding along the first row, the numera- tor is D 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle...
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HW 15 - yang (ey942) ohw15 turner (56705) 1 This print-out...

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