This preview shows pages 1–2. 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: mcconnell (kam2342) homework28 Turner (60230) 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 In an RL series circuit, an inductor of 3 . 58 H and a resistor of 7 . 53 are connected to a 23 . 2 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: 16 . 9918 J. Explanation: Let : L = 3 . 58 H , R = 7 . 53 , and E = 23 . 2 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 = 23 . 2 V 7 . 53 = 3 . 08101 A . The energy stored in the inductor carrying a current 3 . 08101 A is U = 1 2 LI 2 = 1 2 (3 . 58 H) (3 . 08101 A) 2 = 16 . 9918 J . 002 (part 1 of 3) 10.0 points A long solenoid carries a current I 2 . Another coil (of larger diameter than the solenoid) is coaxial with the center of the solenoid, as in the figure below. 2 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 The current I 2 is held constant. The energy stored in the solenoid is given by 1. U = 2 N 2 2 A 1 2 I 2 2 2. U = 1 2 N 2 2 A 2 I 2 2 3. U = 2 A 2 2 I 2 4. U = 2 N 2 2 A 2 2 I 2 5. U = 2 N 2 2 A 2 2 I 2 2 correct 6. U = 2 N 2 A 2 2 I 2 2 7. U = 2 N 2 2 A 2 I 2 2 8. U = 2 N 2 2 2 A 2 I 2 2 Explanation: The magnetic energy density is given by B 2 2 . Inside the solenoid the magnetic field is B = N 2 I 2 2 , and the volume enclosed by the solenoid is...
View Full
Document
 Spring '10
 Turner

Click to edit the document details