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ohw15-solutions - Reilly(mar3978 ohw15 turner(56725 This...

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Reilly (mar3978) – ohw15 – turner – (56725) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A charged particle is projected with its initial velocity parallel to a uniForm magnetic feld. What is the resulting path? 1. straight line parallel to the feld. correct 2. spiral. 3. straight line perpendicular to the feld. 4. circular arc. 5. parabolic arc. Explanation: The Force on a moving charge due to a magnetic feld is given by v F = qvV × v B . IF vV and v B are parallel, then × v B = 0 . Hence the Force on the particle is zero, and the particle continues to move in a straight line parallel to the feld. 002 10.0 points In the diagram below, the resistances are 2 . 6 Ω and 1 . 57 Ω. The current through R 1 is 4 . 36 A. A B R 1 2 R ±ind the potential di²erence between points A and B . Correct answer: 18 . 1812 V. Explanation: Let : R 1 = 2 . 6 Ω , R 2 = 1 . 57 Ω , and I = 4 . 36 A . Currents through elements in a series combi- nation are the same. The equivalent resis- tance is R = R 1 + R 2 , so ± = I R = I ( R 1 + R 2 ) = 4 . 36 A (2 . 6 Ω + 1 . 57 Ω) = 18 . 1812 V . 003 10.0 points The fgure 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 X in A correct 2. Bulb X in B 3. Bulb Y in B 4. Bulb Y in A 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.
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Reilly (mar3978) – ohw15 – turner – (56725) 2 In B , a series circuit, the current is the same through both bulbs. 004 (part 1 of 3) 10.0 points 5 . 6 V 1 . 6 V 4 . 3 V I 1 1 . 8 Ω 2 . 3 Ω I 2 7 . 7 Ω I 3 8 . 2 Ω Find the current I 1 in the 1 . 8 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 0 . 617315 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) Kirchho±’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchho±’s law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 1 . 8 Ω , R B = 2 . 3 Ω , R C = 7 . 7 Ω , R D = 8 . 2 Ω , E 1 = 5 . 6 V , E 2 = 1 . 6 V , and E 3 = 4 . 3 V . Using determinants, I 1 = v v v v v v 0 1 - 1 E 1 + E 2 0 R D E 3 R C R D v v v v v v v v v v v v 1 1 - 1 R A + R B 0 R D 0 R C R D v v v v v v Expanding along the ²rst row, the numera- tor is D 1 = v v v v v v 0 1 - 1 E 1 + E 2 0 R D E 3 R C R D v v v v v v = 0 - 1 v v v v E 1 + E 2 R D E 3 R D v v v v + ( - 1) v v v v E 1
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