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solutions 15

# solutions 15 - Patel Kinal Homework 15 Due 7:00 pm Inst...

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Patel, Kinal – Homework 15 – Due: Mar 28 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Assume that the length of the solenoid is much larger than the solenoid’s radius and that the core of the solenoid is air. Calculate the inductance of a uniformly wound solenoid having 260 turns if the length of the solenoid is 31 cm and its cross-sectional area is 4 cm 2 . Correct answer: 0 . 109611 mH. Explanation: Let : N = 260 , μ 0 = 1 . 25664 × 10 - 6 N / A 2 , = 31 cm = 0 . 31 m , and A = 4 cm 2 = 0 . 0004 m 2 . In this case, we can take the interior magnetic field to be uniform and given by the equation B = μ 0 n I = μ 0 N I , where n is the number of turns per unit length, N . The magnetic flux through each turn is Φ B = B A = μ 0 N A I , where A is the cross-sectional area of the solenoid. Using this expression and the equa- tion for the inductance, L = N Φ B I , we find that L = μ 0 N 2 A = (1 . 25664 × 10 - 6 N / A 2 ) × (260) 2 (0 . 0004 m 2 ) 0 . 31 m = 0 . 000109611 H = 0 . 109611 mH . 002 (part 2 of 2) 10 points Calculate the self-induced emf in the solenoid described in the first part if the current through it is decreasing at the rate of 66 A / s. Correct answer: 7 . 23434 mV. Explanation: Using the equation E = - N d Φ B d t , and given that d I d t = - 66 A / s , we get E = - L d I d t = - (0 . 000109611 H ) ( - 66 A / s ) = 0 . 00723434 V = 7 . 23434 mV . keywords: 003 (part 1 of 1) 10 points An automobile starter motor draws a current of 4 A from a 9 . 7 V battery when operating at normal speed. A broken pulley locks the ar- mature in position, and the current increases to 24 . 8 A. What was the back emf of the motor when operating normally? Correct answer: 8 . 13548 V. Explanation: Let : I = 24 . 8 A , I 0 = 4 A , and E = 9 . 7 V . When not rotating, E = I R , and from this, R = E I = 9 . 7 V 24 . 8 A = 0 . 391129 Ω ,

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Patel, Kinal – Homework 15 – Due: Mar 28 2008, 7:00 pm – Inst: Weathers 2 When rotating, E - E back = I 0 R , or E back = E - I 0 R = 9 . 7 V - (4 A) (0 . 391129 Ω) = 8 . 13548 V . keywords: 004 (part 1 of 3) 10 points An inductor and a resistor are connected with a double pole switch to a battery as shown in the figure. The switch has been in position b for a long period of time. 137 mH 5 . 85 Ω 6 . 4 V S b a If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches 89 mA? Correct answer: 1 . 98713 ms. Explanation: Let : R = 5 . 85 Ω , L = 137 mH , and E = 6 . 4 V . L R E S b a The time constant of an RL circuit is τ = L R = 0 . 137 H 5 . 85 Ω = 0 . 0234188 s . The final current reached in the circuit is I 0 = E R = 6 . 4 V 5 . 85 Ω = 1 . 09402 A . The switch is in position a in an RL circuit connected to a battery at t = 0 when I = 0. Then the current vs. time is I = I 0 1 - e - t / τ · . Solving the above expression for t , when I = I 1 gives t 1 = - τ ln 1 - I 1 I 0 = - (0 . 0234188 s) ln 1 - 0 . 089 A 1 . 09402 A = 1 . 98713 ms . 005 (part 2 of 3) 10 points What is the maximum current in the inductor a long time after the switch is in position a ?
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solutions 15 - Patel Kinal Homework 15 Due 7:00 pm Inst...

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