MP_Assignment#5_soln

MP_Assignment#5_soln - 3/20/11 1:37 PM Assignments Page 1...

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Unformatted text preview: 3/20/11 1:37 PM Assignments Page 1 of 15 http://session.masteringphysics.com/myct/assignments MP_Assignment#5 Due: 8:01am on Monday, March 7, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy Conceptual Question 36.7 Description: (a) In the series RLC circuit represented by the phasors of the figure, is the emf frequency less than, equal to, or greater than the resonance frequency omega_0? (b) Explain. Part A In the series circuit represented by the phasors of the figure, is the emf frequency less than, equal to, or greater than the resonance frequency ? ANSWER: less than equal to greater than Part B Explain. Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: Answer Key: Less than. Here the current leads the emf. X(L) < X(C) and the phase angle phi is negative. Edit Assignment Settings per Student Grade Essays Student View Summary View Diagnostics View Print View with Answers [ Print ] 3/20/11 1:37 PM Assignments Page 2 of 15 http://session.masteringphysics.com/myct/assignments Problem 36.68 Description: (a) Show that the average power loss in a series RLC circuit is P_avg = (( omega^2 (EMF)_rms^2 R)/( omega^2 R^2 + L^2 ( ( omega^2 - omega_0^2 ) )^2 ))... (b) Prove that the energy dissipation is a maximum at omega = omega_0 (kern 1pt). Part A Show that the average power loss in a series circuit is Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: Answer Key: The average power is P(avg)=EMF(avg)*I(avg)*cos(phi), where I(avg)=EMF(avg)/Z and cos(phi)=R/Z. We also have Z^2 = R^2+(omega*L-1/(omega*C))^2 = R^2+L^2/omega^2*(omega^2- omega(res)^2)^2 Combining these results we obtain P(avg) = EMF(avg)*omega^2*R/(R^2*omega^2+L^2*(omega^2- omega(res)^2)^2) Part B Prove that the energy dissipation is a maximum at . Essay answers are limited to about 500 words (3800 characters maximum, including spaces). ANSWER: Answer Key: Energy dissipation will be a maximum, when dP(avg)/domega=0. Taking the derivative, dP(avg)/domega = 2*omega/(omega^2*R^2+L^2*(omega^2-omega(res)^2)^2)-[omega^2*(2*omega*R^2+2*L^2*(omega^2- omega(res)^2)*(2*omega)]/(omega^2*R^2+L^2*(omega^2- omega(res)^2)^2)^2 = 0 => omega^2*R^2+L^2*(omega^2-omega(res)^2)^2 = omega^2*R^2+2*L^2*(omega^2-omega(res)^2)*omega^2 => omega^4 = omega(res)^4 => omega = omega(res). A Series L-R-C Circuit: The Phasor Approach Description: Derive the impedance formula using phasor diagrams and then analyze the formula to obtain the resonance conditions. Learning Goal: To understand the use of phasor diagrams in calculating the impedance and resonance conditions in a series L-R-C circuit....
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This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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MP_Assignment#5_soln - 3/20/11 1:37 PM Assignments Page 1...

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