This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY2049 Fall 2010 Profs. S. Hershfield, A. Petkova Exam 2 Solution 1. Each of the capacitors in the circuit at right has capacitance 1 μ F. What is the effective capacitance of the network? Answer: 0.6 μ F Solution: The effective capacitance of the bottom two capacitors in series is 0.5 μF . These two are in parallel with the top capacitor yielding an effective capacitance of 1.5 μF . Finally, the 1.5 μF capacitor is in series with the 1 μF capacitor on the right, giving a net capacitors of 0.6 μF . 2. If the voltage between points A and B in the circuit at right is 12 volts, what is the charge on the 2 μ F capacitor? Answer: 6 μ C Solution: The three capacitors in series have an effective capacitance of 12 μF . Placing them in series with the 4 μF capacitor on the left yields a net capacitance of 3 μF . When capacitors are in series, the charge is the same. The charge here is (3 μF )(12 V ) = 36 μC . This is the charge on the 3 μF capacitor, which has a voltage drop of (36 μC ) / (3 μF ) = 9 V , and the charge on the 12 μF effective capacitor, which has a voltage drop of (36 μC ) / (12 μF ) = 3 V . When capacitors are in parallel, they have the same voltage. Thus, the charge on the 2 μF capacitor is (2 μF )(3 V ) = 6 μC . 3. A parallel plate capacitor with area 10 cm 2 and plate separation 1 mm is filled with two different dielectrics as shown in the figure. The top half has dielectric constant κ 1 = 10, and the bottom half has dielectric constant κ 2 = 20. What is the capacitance of this parallel plate capacitor? Answer: 1 . 2 × 10 10 F Solution: As in the homework, this capacitor may be viewed as two capacitors in series: a top one with dielectric constant κ 1 and a bottom one with dielectric constant κ 2 . Using C top = κ 1 ǫ o (10 × 10 4 m 2 ) / (0 . 5 × 10 3 m ) and C bottom = κ 1 ǫ o (10 × 10 4 m 2 ) / (0 . 5 × 10 3 m ), the net capacitance is obtained by adding C top and C bottom in series. 4. How are the three circuits at right related? Answer: A and C are equivalent, but not B. Solution: Circuits A and C each have three capacitors in parallel, but circuit B has three capacitors connected in a triangle (not parallel) with one of the capacitors short circuited. 5. A network of wires is shown at right. The currents in some of the wires are labeled. Based on this information, what is the current labeled i? Answer: 1 A Solution: Using “current in = current out”, there is a current of 1 A flowing down the top half of the middle wire. This 1 A combines with the 2 A to give the 3 A current, leaving zero current in the bottom half of the middle wire. Thus, i is equal to 1 A ....
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