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HW04-solutions

# HW04-solutions - Munoz(gm7794 HW04 TSOI(92515 This...

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Munoz (gm7794) – HW04 – TSOI – (92515) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider the circuit 100 V 2 μ F 4 μ F 3 μ F 5 μ F a b c What is the equivalent capacitance for this network? 1. C equivalent = 10 7 μ F 2. C equivalent = 7 3 μ F 3. C equivalent = 14 μ F 4. C equivalent = 7 μ F correct 5. C equivalent = 3 2 μ F Explanation: E B C 1 C 2 C 3 C 4 a b c Let : C 1 = 2 μ F , C 2 = 4 μ F , C 3 = 3 μ F , C 4 = 5 μ F , and E B = 100 V . The equivalent capacitance of capacitors C 1 and C 2 (parallel) is C 12 = C 1 + C 2 = 6 μ F . C 12 and C 3 are in series, so 1 C 123 = 1 C 12 + 1 C 3 = C 3 + C 12 C 12 C 3 C 123 = C 12 C 3 C 3 + C 12 = (6 μ F) (3 μ F) 6 μ F + 3 μ F = 2 μ F . C 123 and C 4 are parallel, so C = C 4 + C 123 = 7 μ F . 002 (part 2 of 2) 10.0 points What is the charge stored in the 5- μ F lower- right capacitor? 1. Q 1 = 1 , 100 μ C 2. Q 1 = 1 , 800 μ C 3. Q 1 = 500 μ C correct 4. Q 1 = 360 μ C 5. Q 1 = 710 μ C Explanation: Let : C 4 = 5 μ F and E B = 100 V . The charge stored in a capacitor is given by Q = C V , so, Q 4 = C 4 V = (5 μ F) (100 V) = 500 μ C . 003 10.0 points Consider the capacitor circuit

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Munoz (gm7794) – HW04 – TSOI – (92515) 2 E B 6 μ F 2 μ F 8 μ F 3 μ F b a c Calculate the equivalent capacitance C ab between points a and b . Correct answer: 4 . 63158 μ F. Explanation: Let : C 1 = 6 μ F , C 2 = 8 μ F , C 3 = 2 μ F , and C 4 = 3 μ F . E B C 1 C 3 C 2 C 4 b a c For capacitors in series, 1 C series = 1 C i V series = V i , and the individual charges are the same. For parallel capacitors, C parallel = C i Q parallel = Q i , and the individual voltages are the same. Let c be the point between the two parallel sections. Let c be the point between the two parallel sections. The capacitance between a an c is C l = C 1 + C 3 = 6 μ F + 2 μ F = 8 μ F . The capacitance between b and c is C r = C 2 + C 4 = 8 μ F + 3 μ F = 11 μ F . Thus the capacitance between a and b is 1 C ab = 1 C l + 1 C r = C r + C l C l C r C ab = C l C r C l + C r = (8 μ F) (11 μ F) 8 μ F + 11 μ F = 4 . 63158 μ F . 004 (part 1 of 2) 10.0 points Consider two conductors 1 and 2 made of the same ohmic material; i.e. , ρ 1 = ρ 2 . Denote the length by , the cross sectional area by A . The same voltage V is applied across the ends of both conductors. The field E is inside of the conductor. V 1 E 1 I 1 1 r 1 V 2 E 2 I 2 2 r 2 If A 2 = 2 A 1 , 2 = 2 1 and V 2 = V 1 , find the ratio E 2 E 1 of the electric fields.
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