QTest #5-solutions - Hint: You may fnd it helpFul to work...

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Version 089/ABBCB – QTest #5 – Antoniewicz – (56495) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points An 11 Ω resistor and a 4.0 Ω resistor are connected in series with an emf source. The potential di±erence across the 4.0 Ω resistor is measured with a voltmeter to be 18 V. ²ind the potential di±erence across the emf source. 1. 63.0 2. 84.0 3. 48.0 4. 44.0 5. 67.5 6. 96.0 7. 16.0 8. 72.0 9. 180.0 10. 81.0 Correct answer: 67 . 5 V. Explanation: Let : R 1 = 11 Ω , R 2 = 4 . 0 Ω , and Δ V 2 = 18 V . Basic Concepts: R eq = R 1 + R 2 Δ V = IR I 1 = I 2 = I Solution: I = Δ V 2 R 2 = 18 V 4 Ω = 4 . 5 A R eq = 11 Ω + 4 Ω = 15 Ω Δ V = IR eq = (4 . 5 A)(15 Ω) = 67 . 5 V . 002 10.0 points E A B C D E Rank the brightness oF the identical bulbs in the Following circuit.
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Unformatted text preview: Hint: You may fnd it helpFul to work out the currents through the bulbs For the case V = 1 V , and R = 1 For all the bulbs, then compare the currents. 1. A = B = C = D = E 2. E = D > A > B = C 3. A = B > C > D = E 4. A = D = E > B > C 5. B = C > A = D = E 6. A = B = C > D > E 7. C > B > A > D > E 8. A = C > B > D = E 9. A = B = C > D = E 10. A = D = E > B = C correct Explanation: rom the fgure, with all the bulbs having the resistance R , I A = I D = I E = V R I B = I C = V 2 R Version 089/ABBCB QTest #5 Antoniewicz (56495) 2 The rankings of the currents shown here are also the corresponding rankings of the bright-ness....
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QTest #5-solutions - Hint: You may fnd it helpFul to work...

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