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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 ....
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 Fall '08
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 Physics, Capacitance, Work, Magnetic Field, Electric charge

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