Exam 2_20230203 Solutions

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Electrical Engineering

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Feb 20, 2024

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PHYS 102 Exam 2 Spring 2023-02-16 Two parts worth a total of 25 points: multiple choice (4 questions; 4 points) and written answer (2 questions; 11 and 10 points respectively). Multiple-choice answers are to be written on the bubble sheet at the end of the test. All multiple- choice questions have only one correct answer. You may detach the bubble sheet when the test starts. Written answer questions must be answered on the page(s) following the question. Partial marks will be awarded for solutions that are partially correct. Solutions Exam advice: • Write your name and student number on the front page. • Read or skim the exam to the end before starting. • Do the easier questions first. • Draw diagrams even if it’s not demanded by the question. • Show your work and write explanations neatly. Use full sentences! • Include units and reasonable significant figures for all numerical answers and directions for all vector quantities. • Put a box around your final answer in all cases. Possibly-useful facts: • Permittivity of free space: • Coulomb’s constant:

1. A flashlight with a bulb of resistance 50 Ω is powered by a 4.5 V battery. Which of the following changes would increase the brightness of the bulb? A) Add a second identical bulb to the circuit in parallel with the existing light bulb. B) Add a second identical bulb to the circuit in series with the existing light bulb. C) Add a second identical battery to the circuit in parallel with the existing battery. D) Add a second identical battery to the circuit in series with the existing battery. E) More than one of these is correct. A) A second bulb in parallel would increase the current draw from the battery, but this current would be split between the batteries, leaving the brightness unchanged. B) A second bulb in series would decrease the current through both bulbs, decreasing the brightness. C) A second battery in parallel would not change the current or voltage through the bulb. Each battery would supply half as much current. D) A second battery in series would double the potential difference across the bulb. According to Ohm’s law that would double the current, which increases the brightness.

2. A capacitor with capacitance C 1 is fully charged through a resistor with resistance R 1 using a battery with potential V bat = 12 V. A total of 100 mJ of energy is stored on the capacitor in this manner. The battery is then removed and replaced with a device that behaves like a resistor of resistance R L = 1 kΩ. After 1.0 seconds, the capacitor has discharged half of its energy . What is the capacitance C 1 and the resistance R 1 ? A) C 1 = 1.38 mF R 1 = 2077 Ω B) C 1 = 720 F R 1 = 2077 Ω C) C 1 = 1.38 mF R 1 = 3784 Ω D) C 1 = 720 F R 1 = 1077 Ω E) C 1 = 1.38 mF R 1 = 1077 Ω To find the capacitance, use the relation U = ½CV 2 reaaranged to solve for C. C 1 = 2U/V 2 , plugging in the values given for U and V we get C 1 = 1.39 mF. (there is a rounding error in the options provided) The voltage on a discharging capacitor follows the function V(t) = V 0 exp(-t/τ). In our case V 0 = V bat and τ = (R 1 +R L )C. Using the same relation as above, we can write the function for energy: U(t) = ½C (V bat exp(-t/τ)) 2 = ½CV bat 2 exp(-2t/τ) We are told that U(t 1 ) = U(0)/2 where t 1 is 1.0s, so we can write: U(0)/2 = ½½CV bat 2 exp(0) = ½CV bat 2 exp(-2t 1 /τ) which simplifies to: ½ = exp(-2t 1 /τ). Solving for τ and substituting the resistors, we obtain R 1 = 2t 1 /(C ln 2) - R L. Putting in the values given we get R1 = 1077 Ω .

3. In which direction is the force exerted on a negative charge in the following uniform magnetic field? A) X B) ⊙ C) At a different angle, in the plane of the page D) At a different angle, out of the plane of the page E) F = 0 This is a negative charge, so we either use the right-hand-rule and reverse the direction, or use our left-hand. The result is out of the page. The velocity and the magnetic field are perpendicular to each other, so the force is not zero. The force is always perpendicular to both the magnetic field and velocity, so it had to be directly into or out of the page, not at some other angle.

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