Heimadæmi 9 Due: 11:00pm on Thursday, March 17, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 31.48 Figure shows a lowpass filter; the output voltage is taken across the capacitor in an LRC series circuit. Part A Derive an expression for Vout /Vs, the ratio of the output and source voltage amplitudes, as a function of the angular frequency ω of the source. ANSWER: V out Vs V out Vs V out Vs V out Vs 1 = 2 ω C √R +(ω L− 1 2 ) ω C 1 = 2 ω C √R −(ω L− 1 2 ) ω C 1 = 2 ω C √R −(ω L+ 1 2 ) ω C 1 = 2 ω C √R +(ω L+ 1 2 ) ω C Correct Part B Find an expression for Vout /Vs when ω is large. ANSWER: 1 1 2 (L C) ω 1 (L C) ω (L C) ω2 Correct Part C Find an expression for Vout /Vs in the limit of small frequency. ANSWER: 1 1 2 (L C) ω 1 (L C) ω (L C) ω2 Correct Problem 31.55 −−−−−−−−−−−−−−−−−− The impedance of an LRC parallel circuit can be written as 1/Z 2 2 = √ (1/R ) + [ωC − 1/ωL] . Part A Show that at the resonance angular frequency ω0 current through the ac source is a minimum. ANSWER: −− − = 1/√LC , the impedance Z is a maximum and therefore the Using −−−−−−−−−−−−−−−−−− 2 2 1/Z = √ (1/R ) + [ωC − 1/ωL] we see that the impedance is a maximum when the square Z root is a minimum, and that occurs when ωC − 1/ωL resonance angular frequency ω0 −− − = 1/√LC . Since I . But that occurs when ω = 0 = −− − = 1/√LC which is the V /Z, the current is then a minimum when Z is a maximum. Using −−−−−−−−−−−−−−−−−− 2 2 1/Z = √ (1/R ) + [ωC − 1/ωL] we see that the impedance is a maximum when the square Z root is a maximum, and that occurs when ωC − 1/ωL resonance angular frequency ω0 −− − = 1/√LC . Since I . But that occurs when ω = 0 = V /Z −− − = 1/√LC which is the , the current is then a minimum when Z is a maximum. Using −−−−−−−−−−−−−−−−−− 2 2 1/Z = √ (1/R ) + [ωC − 1/ωL] we see that the impedance is a maximum when the square Z root is a minimum, and that occurs when ωC − 1/ωL resonance angular frequency ω0 −− − = 1/√LC . Since I . But that occurs when ω = 0 = Z/V −− − = 1/√LC which is the , the current is then a minimum when Z is a maximum. Using −−−−−−−−−−−−−−−−−− 2 2 1/Z = √ (1/R ) + [ωC − 1/ωL] we see that the impedance is a maximum when the square Z root is a maximum, and that occurs when ωC − 1/ωL resonance angular frequency ω0 −− − = 1/√LC . Since I . But that occurs when ω = 0 = Z/V −− − = 1/√LC which is the , the current is then a minimum when Z is a maximum. Correct Part B A 100Ω resistor, a 0.100μF capacitor, and a 0.300H inductor are connected in parallel to a voltage source with amplitude 240 V.What is the resonance angular frequency? ANSWER: ω0 = 5.77×103 rad/s Correct Part C For the circuit that was described in part B find the maximum current through the source at the resonance frequency. Express your answer to three significant figures and include the appropriate units. ANSWER: I = 2.40 A Correct Part D For the circuit that was described in part B find the maximum current in the resistor at resonance. Express your answer to three significant figures and include the appropriate units. ANSWER: IR = 2.40 A Correct Part E For the circuit that was described in part B find the maximum current in the inductor at resonance. Express your answer to three significant figures and include the appropriate units. ANSWER: IL = 0.139 A Correct Part F For the circuit that was described in part B find the maximum current in the branch containing the capacitor at resonance. Express your answer to three significant figures and include the appropriate units. ANSWER: IC = 0.139 A Correct Problem 31.51 In an LRC series circuit the magnitude of the phase angle is 54.2 ∘ with the source voltage lagging the current. The reactance of the capacitor is 360 Ω, and the resistor resistance is 174 Ω. The average power delivered by the source is 133 W . Part A Find the reactance of the inductor. ANSWER: XL = 119 Ω Correct Part B Find the rms current. ANSWER: Irms = 0.874 A Correct Part C Find the rms voltage of the source. ANSWER: Vrms = 260 V Correct Exercise 31.3 The voltage across the terminals of an ac power supply varies with time according to V amplitude is V0 = 43.0 V . = V0 cos(ωt) . The voltage Part A What is the rootmeansquare potential difference Vrms ? ANSWER: Vrms = 30.4 V Correct Part B What is the average potential difference Vav between the two terminals of the power supply? ANSWER: Vav = 0 V Correct Exercise 31.1 You have a special lightbulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.30 A , even for an instant. Part A What is the largest rootmeansquare current you can run through this bulb? ANSWER: Irms = 0.919 A Correct Exercise 31.7 The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 179 V and frequency 56.0 Hz applied across the capacitor is to produce a current amplitude of 0.854 A through the capacitor. Part A What capacitance C is required? ANSWER: C = 1.36×10−5 F Correct Score Summary: Your score on this assignment is 99.1%. You received 5.94 out of a possible total of 6 points.

Problem 31.48