This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY2049 Spring 2010 Profs. P. Avery, A. Rinzler, S. Hershfield Exam 2 Solution 1. In the Figure the capacitances are in microFarads ( 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 crosssection. 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 currentin is equal to currentout. The current in the wider section is 14 C/ (30 s ) = 467 mA . 5. A copper wire has a length L , a square crosssection 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 crosssections (see Figure). These are then welded together endtoend 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....
View
Full
Document
This note was uploaded on 10/26/2011 for the course PHY 2049 taught by Professor Any during the Fall '08 term at University of Florida.
 Fall '08
 Any
 Physics, Capacitance, Work

Click to edit the document details