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HW 6 - segovia(cs39966 Homework 6 Spurlock(14441 This...

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segovia (cs39966) – Homework 6 – Spurlock – (14441) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points How much energy is stored in a 73 mH induc- tor at the instant when the current is 3 . 2 A? Correct answer: 0 . 37376 J. Explanation: Let : L = 73 mH = 0 . 073 H and I = 3 . 2 A . The energy is U = 1 2 L I 2 = 1 2 (0 . 073 H)(3 . 2 A) 2 = 0 . 37376 J . 002 10.0points An inductor of 240 turns has a radius of 7 cm and a length of 29 cm. The permeability of free space is 1 . 25664 × 10 - 6 N / A 2 . Find the energy stored in it when the cur- rent is 0 . 6 A. Correct answer: 0 . 000691596 J. Explanation: Let : r = 7 cm = 0 . 07 m , = 29 cm = 0 . 29 m , N = 240 turns , I = 0 . 6 A , and μ 0 = 1 . 25664 × 10 - 6 N / A 2 . The inductance is L = μ 0 N 2 A = μ 0 N 2 π r 2 = (1 . 25664 × 10 - 6 N / A 2 ) × (240 turns) 2 π (0 . 07 m) 2 0 . 29 m = 0 . 0038422 H . The energy stored in an inductor is U = 1 2 L I 2 = 1 2 (0 . 0038422 H)(0 . 6 A) 2 = 0 . 000691596 J . 003 10.0points A “slinky” toy spring has a radius of 3 . 4 cm and an inductance of 330 μ H when extended to a length of 0 . 6 m. How many turns are in the spring? Correct answer: 208 . 293 turns. Explanation: Let : r = 3 . 4 cm = 0 . 034 m , L = 330 μ H = 0 . 00033 H , and = 0 . 6 m . From L = μ 0 N 2 A , we find N = radicalBigg L ℓ μ 0 A = radicalBigg (0 . 00033 H)(0 . 6 m) (1 . 25664 × 10 - 6 n / A 2 ) π (0 . 034 m) 2 = 208 . 293 turns . 004(part1of3)10.0points 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. 200 mH 6 . 13 Ω 7 . 6 V S b a
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segovia (cs39966) – Homework 6 – Spurlock – (14441) 2 If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches 155 mA? Correct answer: 4 . 35739 ms. Explanation: Let : R = 6 . 13 Ω , L = 200 mH , and E = 7 . 6 V . L R E S b a The time constant of an RL circuit is τ = L R = 0 . 2 H 6 . 13 Ω = 0 . 0326264 s . The final current reached in the circuit is I 0 = E R = 7 . 6 V 6 . 13 Ω = 1 . 2398 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 parenleftBig 1 - e - t / τ parenrightBig . Solving the above expression for t , when I = I 1 gives t 1 = - τ ln parenleftbigg 1 - I 1 I 0 parenrightbigg = - (0 . 0326264 s) ln parenleftbigg 1 - 0 . 155 A 1 . 2398 A parenrightbigg = 4 . 35739 ms . 005(part2of3)10.0points What is the maximum current in the inductor a long time after the switch is in position a ? Correct answer: 1 . 2398 A. Explanation: After a long time compared to τ , we have a d.c. circuit with a battery supplying an emf E , which is equal to the voltage drop I R across the resistor. Thus I = E R = 7 . 6 V 6 . 13 Ω = 1 . 2398 A .
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