HW10-Solutions - Version PREVIEW HW 10 feng (58230) This...

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Version PREVIEW – HW 10 – feng – (58230) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. Tipler PSE5 29 119 001 (part 1 of 4) 10.0 points In the circuit shown in Fgure, E 1 = (32 V) cos 2 πft , f = 210 Hz, E 2 = 6 V, and R = 32 Ω. E 1 = E max cos ωt E max = 32 V E 2 R ±ind the maximum current through the re- sistor. Correct answer: 1 . 1875 A. Explanation: Let : R = 32 Ω , E 1 = (32 V) cos 2 π f t , and E 2 = 6 V . Applying Kirchho²’s loop rule, we have E 1 + E 2 - I R = 0 , so the current is I ( t ) = E 1 + E 2 R = I 1 cos 2 π f t + I 2 , where I 1 = 32 V 32 Ω = 1 A , and I 2 = 6 V 32 Ω = 0 . 1875 A , since the maximum current occurs when cos 2 π f t = 1 . Thus the maximum current is I max = I 1 + I 2 = 1 A + 0 . 1875 A = 1 . 1875 A . 002 (part 2 of 4) 10.0 points ±ind the minimum current through the resis- tor. Correct answer: - 0 . 8125 A. Explanation: The minimum current occurs at cos 2 π f t = - 1 , so the maximum value is I max = - I 1 + I 2 = - (1 A) + 0 . 1875 A = - 0 . 8125 A . 003 (part 3 of 4) 10.0 points ±ind the average current through the resistor. Correct answer: 0 . 1875 A. Explanation: Because the average of cos ω t is zero, I av = I 2 = 0 . 1875 A . 004 (part 4 of 4) 10.0 points ±ind the rms current through the resistor. Correct answer: 0 . 731544 A. Explanation: Let : f = 210 Hz , and I 1 ( t ) = (1 A) cos 2 π f t . The square of I ( t ) is I 2 ( t ) = ( I 1 ( t ) + I 2 ) 2 = I 2 1 ( t ) + 2 I 1 ( t ) I 2 + I 2 2 = I 2 1 cos 2 2 π f t + 2 I 1 I 2 cos 2 π f t + I 2 2
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Version PREVIEW – HW 10 – feng – (58230) 2 The rms current is obtained by integrating I 2 ( t ) from t = 0 to t = T = 1 f and then dividing it by T . Because 1 T i T 0 cos 2 2 π f t dt = 1 2 T i T 0 (1 + 2 cos 2 π f t ) dt = 1 2 T T = 1 2 and the average of cos ω t is zero, we have I 2 av = 1 2 I 2 1 + I 2 2 = 1 2 (1 A) 2 + (0 . 1875 A) 2 = 0 . 535156 A 2 . Thus the rms current is I rms = r I 2 av = 0 . 535156 A 2 = 0 . 731544 A . Frequency and Current 005 (part 1 of 2) 10.0 points In the circuit below, the inductor is 183 mH, the maximum voltage is 71 . 8 V, and the in- ductive reactance is 40 Ω. E = E max sin ωt E max = 71 . 8 V S 183 mH What is the current’s frequency in the in- ductor? Correct answer: 34 . 788 Hz. Explanation: Let : X L = 40 Ω , and L = 183 mH = 0 . 183 H . The frequency of the current is the same as that of the electric potential in the circuit. The inductive reactance is given by X L = ω L , so ω = X L L and the frequency in the inductor is f = ω 2 π = X L 2 π L = 40 Ω 2 π (0 . 183 H) = 34 . 788 Hz . 006 (part 2 of 2) 10.0 points Calculate the maximum current at this fre- quency.
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HW10-Solutions - Version PREVIEW HW 10 feng (58230) This...

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