This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: choe (dec865) Homework 17 Weathers (22202) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points An automobile starter motor draws a current of 2 . 3 A from a 18 . 1 V battery when operating at normal speed. A broken pulley locks the ar mature in position, and the current increases to 14 . 1 A. What was the back emf of the motor when operating normally? Correct answer: 15 . 1475 V. Explanation: Let : I = 14 . 1 A , I = 2 . 3 A , and E = 18 . 1 V . When not rotating, E = I R , and from this, R = E I = 18 . 1 V 14 . 1 A = 1 . 28369 , When rotating, E E back = I R , or E back = E I R = 18 . 1 V (2 . 3 A) (1 . 28369 ) = 15 . 1475 V . 002 10.0 points In an RL series circuit, an inductor of 4 . 43 H and a resistor of 6 . 49 are connected to a 21 . 4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Correct answer: 24 . 0831 J. Explanation: Let : L = 4 . 43 H , R = 6 . 49 , and E = 21 . 4 V . The current in an RL circuit is I = E R parenleftBig 1 e Rt/L parenrightBig . The final equilibrium value of the current, which occurs as t , is I = E R = 21 . 4 V 6 . 49 = 3 . 29738 A . The energy stored in the inductor carrying a current 3 . 29738 A is U = 1 2 L I 2 = 1 2 (4 . 43 H) (3 . 29738 A) 2 = 24 . 0831 J . 003 10.0 points A uniform electric field of magnitude 1 . 02 10 6 V / m throughout a cylindrical volume re sults in a total energy of 1 . 53 J. What magnetic field over this same region stores the same total energy? Correct answer: 0 . 00340235 T. Explanation: Let : E = 1 . 02 10 6 V / m . The electric field density is u E = E 2 2 . The magnetic field density is u B = B 2 2 . We are looking for B such that U B = U M u B V = u E V . choe (dec865) Homework 17 Weathers (22202) 2 In the above equation, V is the volume of the region under consideration. The speed of light c = 1 , so B 2 = E 2 B = E = E c = 1 . 02 10 6 V / m 2 . 99792 10 8 m / s = . 00340235 T . 004 (part 1 of 3) 10.0 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. 214 mH 7 . 44 7 . 9 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 148 mA? Correct answer: 4 . 31753 ms. Explanation: Let : R = 7 . 44 , L = 214 mH , and E = 7 . 9 V ....
View
Full
Document
This document was uploaded on 09/23/2010.
 Spring '09
 Current, Work

Click to edit the document details