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Unformatted text preview: PHY2049 Fall 2009 Profs. A. Petkova, A. Rinzler, S. Hershfield Exam 2 Solution 1. Three capacitor networks labeled A, B & C are shown in the figure with the individual capacitor values labeled (all units μ F). Connection to each network is to be made via the round black end terminals. Each network has an equivalent capacitance. The equivalent capacitances of the three networks can be rank ordered, greatest first, in the order: Answer: A, B, C Solution: The equivalent capacitances of the three circuits are C A = 12 μF , C B = 9 μF , and C C = 7 μF . 2. The capacitance of a parallelplate capacitor can be increased by: Answer: decreasing the plate separation Solution: The capacitance of a parallelplate capacitor is C = ǫ o A/d . Increasing d causes C to decrease. 3. In the circuit shown all capacitors are 6 . μ F and the power supply is 12V. The charge on the capacitor labeled q is: Answer: 29 μ C Solution: The effective capacitance of the capacitor with q and the two capacitors in parallel with it is 9 μF . Placing the capacitors in series with a 6 μF capacitor gives an effective capacitance of 3.6 μF . The charge on this effective capacitor is (3 . 6 μF )(12 V ) = 43 . 2 μC . This is also the charge on the 9 μF effective capacitor so the voltage across it is 42 . 3 μC/ 9 μF = 4 . 8 V . Because capacitors in parallel have the same voltage, the voltage across the q capacitor is 4.8V, and q = (6 μF )(4 . 8 V ) = 28 . 8 μC . 4. A thin, flat, square sheet of metal having an edge length of 2 m has one large flat side coated with a uniform 0.10 mm thick layer of strontium titanate (dielectric constant κ = 310). A second identical metal plate lies atop the titanate layer symmetrically sandwiching it between the two plates forming a capacitor. Highly conducting metal wires couple electrically to each plate. Using a 12 V power supply, you fully charge the capacitor followed by disconnecting it from the power supply and touching the two wires together to fully discharge the capacitor. Ignoring resistive losses, how many times must you charge and discharge the capacitor to draw one Joule of energy from the supply? Answer: 127 Solution: The two sheets with a dielectric in between have capacitance C = κǫ o A/d = 1 . 1 × 10 4 F . The energy stored when the capacitor is placed across a voltage of 12V is U = 1 2 CV 2 = 7 . 9 × 10 3 J . The number of times one would have to charge the capacitor to get 1 J of energy is 1 J/ 7 . 9 × 10 3 J = 127. 5. A wire having a length of 150m, and radius of 0.15mm, has 1V applied across its two ends so that it carries a current with a uniform current density of 2 . 8 × 10 5 A/m 2 . The resistivity of the wire (in units of Ω · m) must be: Answer: 2 . 4 × 10 8 Solution: The resistance of a wire is R = ρL/A so the resistivity is ρ = RA/L =...
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 Spring '08
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 Physics, Work, Magnetic Field, Electric charge, Electrical resistance

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