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HW08-solutions-1

# HW08-solutions-1 - gilbert(amg3448 HW08 tsoi(57210 This...

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gilbert (amg3448) – HW08 – tsoi – (57210) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 5) 10.0 points In a series RLC AC circuit, the resistance is 30 . 6 Ω , the inductance is 33 . 6 mH, and the capacitance is 7 . 22 μ F. The maximum potential is 139 V, and the angular frequency is 636 . 62 rad / s. Calculate the maximum current in the cir- cuit. Correct answer: 0 . 700098 A. Explanation: Let : ω = 636 . 62 rad / s , C = 7 . 22 μ F = 7 . 22 × 10 - 6 F , L = 33 . 6 mH = 0 . 0336 H , R = 30 . 6 Ω , and V max = 139 V . The capacitive reactance is X C = 1 ω C = 1 (636 . 62 rad / s) (7 . 22 × 10 - 6 F) = 217 . 562 Ω , and the inductive reactance is X L = ω L = (636 . 62 rad / s) (0 . 0336 H) = 21 . 3904 Ω . The maximum current is I max = V max Z = V max R 2 + ( X L - X C ) 2 = 139 V (30 . 6 Ω ) 2 + (21 . 3904 Ω - 217 . 562 Ω ) 2 = 0 . 700098 A . 002 (part 2 of 5) 10.0 points Determine the maximum voltage across the resistor. Correct answer: 21 . 423 V. Explanation: The maximum voltage across the resistor is V R = I max R = (0 . 700098 A) (30 . 6 Ω ) = 21 . 423 V . 003 (part 3 of 5) 10.0 points Determine the maximum voltage across the inductor. Correct answer: 14 . 9754 V. Explanation: The maximum voltage across the inductor is V L = I max X L = (0 . 700098 A) (21 . 3904 Ω ) = 14 . 9754 V . 004 (part 4 of 5) 10.0 points Determine the maximum voltage across the capacitor. Correct answer: 152 . 315 V. Explanation: The maximum voltage across the capacitor is V C = I max X C = (0 . 700098 A) (217 . 562 Ω ) = 152 . 315 V . 005 (part 5 of 5) 10.0 points What is the power factor for the circuit? Correct answer: 0 . 154122. Explanation:

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gilbert (amg3448) – HW08 – tsoi – (57210) 2 The phase angle between the voltage and the current in the circuit is φ = tan - 1 X L - X C R . Since X L - X C R = 21 . 3904 Ω - 217 . 562 Ω 30 . 6 Ω = - 6 . 41083 , the power factor is cos φ = cos tan - 1 X L - X C R = cos tan - 1 ( - 6 . 41083) = 0 . 154122 . 006 (part 1 of 2) 10.0 points A resistance R and a 1 . 2 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 25 V and the voltage across the inductor is 45 V. What is the resistance R ? Correct answer: 251 . 327 Ω . Explanation: Let : L = 1 . 2 H , f = 60 Hz , V R = 25 V , and V L = 45 V . The potential di ff erence across an inductor is V L = I X L = I 2 π f L and the potential di ff erence across a resistor is V R = I R . Thus we have V R V L = I R I 2 π f L = R 2 π f L , so the resistance is R = V R V L 2 π f L = 25 V 45 V 2 π (60 Hz) (1 . 2 H) = 251 . 327 Ω . 007 (part 2 of 2) 10.0 points What is the AC input voltage? Correct answer: 51 . 4782 V. Explanation: Because V R leads V L by 90 in an LR circuit, the ac input voltage is V = V 2 R + V 2 L = (25 V) 2 + (45 V) 2 = 51 . 4782 V . 008 10.0 points Calculate the resonance frequency of a se- ries RLC circuit for which the capacitance is 60 μ F , the resistance is 45 k Ω , and the induc- tance is 146 mH . Correct answer: 53 . 7735 Hz. Explanation: Let : R = 45 k Ω = 45000 Ω , L = 146 mH = 0 . 146 H , and C = 60 μ F = 6 × 10 - 5 F . The resonance frequency is the frequency at which the current becomes maximum, or the impedance becomes minimum. This occurs when X L = X C ω L = 1 ω C .
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