exam2aS - Physics 21 Fall, 2007 Solution, Hour Exam #2 The...

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Physics 21 Fall, 2007 Solution, Hour Exam #2 The graders for the problems were: 1 McGeehan, 2 Belony, 3 Sweeney, 4 Lyu, 5 Kanofsky For questions about the grading, see the grader by Nov. 20. Problem 1. For the following circuit, R =5 . 0Ω , L = 40 . 0mH ,and C = 300 nF, and the frequency is 1000 Hz. R L C V rms 5 I rms 530.5 I rms 251.3 I rms 5 I rms 279.2 I rms V rms φ (a) (b) (a) (5 pts.) Draw a phasor diagram for the voltages in this circuit. What is the rms value of the current if V rms = 6V? The phasor diagram is shown in panel (a). The quanti- ties are determined from V R = RI rms =5 I rms V L = ωLI rms =2 π × 1000 × 0 . 040 = 251 . 3 I rms V C = I rms ωC = I rms 2 π × 1000 × 300 × 10 9 = 530 . 5 I rms Then add V L + V C [panel (b)] and solve for V rms = p 5 2 + 279 . 2 2 I rms = 279 . 2 I rms I rms = V rms 279 . 2 =21 . 5mA (b) (5 pts.) What is the phase relationship between the current and the applied ac voltage? In other words, does the current lag or lead the applied voltage? by how much? tan φ = 279 . 2 5 = 55 . 8 φ = 88 . 97 The current leads the voltage by almost 90 . (c) (5 pts.) What is the power delivered to the circuit by the power supply? P av = V rms I rms cos φ =6( . 0215)( . 0179) = 2 . 3 × 10 3 W (d) (5 pts.) At what frequency would the rms current be the maximum for the given V rms ? (It is not necessary to derive this result; just show the expression that you use to calculate it.) ω = 1 LC = 1 p . 040(300 × 10 9 ) =9 . 13 × 10 3 rad / s Problem 2. Consider a straight segment of wire of length a carrying a current I . (The rest of the wire need not be considered and is not shown.) P =( x, 0,0) (0, - a /2,0) (0 ,+a /2,0) x y I d l r - r' r' r (a) (5 pts.) Use the Biot-Savart law to write an expres- sion for the magnetic ±eld d B at the point P on the x axis due to the current I in the in±nitesimal segment of wire d l located at (0
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exam2aS - Physics 21 Fall, 2007 Solution, Hour Exam #2 The...

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