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1178864707Xuj - Alvarado Patrick Homework 6 Due Mar 1 2007...

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Alvarado, Patrick – Homework 6 – Due: Mar 1 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A capacitor network is shown below. 145 V 9 μ F 9 μ F 9 μ F 9 μ F 4 μ F 9 μ F 9 μ F y z What is the equivalent capacitance between points y and z of the entire capacitor net- work? Correct answer: 7 . 27273 μ F. Explanation: Let : C a = C = 9 μ F , C b = C = 9 μ F , C c = C = 9 μ F , C d = C = 9 μ F , C e = C = 9 μ F , C f = C = 9 μ F , C x = 4 μ F = 4 × 10 - 6 F and E B = V = 145 V . E R C a C f C b C d C x C e C c y z For capacitors in series, 1 C series = X 1 C i V series = X V i , and the individual charges are the same. For parallel capacitors, C parallel = X C i Q parallel = X Q i , and the individual voltages are the same. The capacitors C b , C c , and C d are in series, so 1 C bcd = 1 C + 1 C + 1 C = 3 C C bcd = 1 3 C . This reduces the circuit to E R C a C f C x C e C bcd y z The capacitors C e and C bcd are parallel, so C bcde = C + C bcd = C + 1 3 C = 4 3 C . This reduces the circuit to E R C a C f C x C bcde y z The capacitors C a , C bcde and C f are in series, so 1 C abcdef = 1 C + 3 4 C + 1 C = 11 4 C C abcdef = 4 11 C . This reduces the circuit to E R C x C abcdef y z
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Alvarado, Patrick – Homework 6 – Due: Mar 1 2007, 11:00 pm – Inst: Opyrchal, H 2 These capacitors are parallel, so C yz = C x + C abcdef = C x + 4 11 C = 4 μ F + 4 11 (9 μ F) = 7 . 27273 μ F . 002 (part 2 of 2) 10 points What is the charge on the 4 μ F capacitor centered on the left directly between points y and z ? Correct answer: 0 . 00058 C. Explanation: C q V q = C x V = (4 × 10 - 6 F) (145 V) = 0 . 00058 C . keywords: 003 (part 1 of 2) 10 points Four capacitors are connected as shown in the figure. 23 μ F 59 μ F 48 μ F 72 μ F 92 V a b c d Find the capacitance between points a and b . Correct answer: 121 . 467 μ F. Explanation: Let : C 1 = 23 μ F , C 2 = 48 μ F , C 3 = 59 μ F , C 4 = 72 μ F , and E = 92 V .
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