exam2sol - PHY2049 Spring 2010 Profs P Avery A Rinzler S...

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Unformatted text preview: PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield Exam 2 Solution 1. In the Figure the capacitances are in micro-Farads ( μ F). The equivalent capac- itance of the network of capacitors is: Answer: 4.5 μ F Solution: The 6 μF and 12 μF capacitors in parallel have capacitance 18 μF . These capacitors are in series with the other 6 μF capacitor to the effective capacitance is 4 . 5 μF . 2. In the previous problem, the charge on the positive plate of the top 6 μ F capacitor is: Answer: 18 μ C Solution: 12 volts across an effective 4 . 5 μF capacitor gives a charge of CV = 54 μC . The 6 μF and 12 μF capacitors in parallel have effective capacitance of 18 μF so the voltage across them is Q/C = 54 μC/ 18 μF = 3 V . A voltage of 3 V across a 6 μF capacitor corresponds to a charge of (6 μF )(3 V ) = 18 μC . 3. A parallel plate capacitor has a plate area of 45 cm 2 a plate separation of 30 micrometers and is filled with a dielectric material having a dielectric constant of 280. When charged to 6.0 V the potential energy stored in the device is: Answer: 6.7 μJ Solution: The capacitance of the parallel plate capacitor is C = κǫ o A/d . The energy stored in the capacitor is U = (1 / 2) CV 2 = 6 . 7 μJ . 4. A particular wire has a circular cross-section. For the first 1 / 2 meter of its one meter length its radius is 4 mm while for the second half it is 2 mm. 14 Coulombs of charge pass a point in the 4 mm section in 30 seconds. The current in the 2 mm section is: Answer: 467 mA Solution: The current through both sections of the wire is the same because current-in is equal to current-out. The current in the wider section is 14 C/ (30 s ) = 467 mA . 5. A copper wire has a length L , a square cross-section of side length W and an end to end resistance R . Without loss of material the wire is sliced along its long direction to make 4 equal length wires of equal square cross-sections (see Figure). These are then welded together end-to-end to make one wire that is 4 times the original length. The new resistance is: Answer: 16 R Solution: The resistance of a wire is R = ρL/A . If L → 4 L and A → A/ 4, then the resistance increases by a factor of 16. 6. A 24.0V DC motor raises a 1.00 kg mass with a constant speed through 10.0 meters in 20.0 seconds. Assuming no other energy losses the current drawn by the motor must be: Answer: 204 mA Solution: If there are no other energy losses, the power output of the motor is equal to the power input into the motor....
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This note was uploaded on 04/23/2010 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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exam2sol - PHY2049 Spring 2010 Profs P Avery A Rinzler S...

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