Ch23-h3-solutions - pan (zp695) Ch23-h3 chiu (56565) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: pan (zp695) Ch23-h3 chiu (56565) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points One application of an RL circuit is the gen- eration of time-varying high-voltage from a low-voltage source, as shown in the figure. 3 H 10 1238 11 . 2 V S b a What is the current in the circuit a long time after the switch has been in position a ? Correct answer: 1 . 12 A. Explanation: L R 2 R 1 E S b a Let : R 2 = 10 and E = 11 . 2 V . When the switch is at a , the circuit com- prises the battery, the inductor L , and the resistor R 2 . A long time after the switch has been in position a , the current is steady. This means that the inductor has no response to the current. In other words, the circuit can be considered as consisting of E and R 2 only, with the inductor reduced to a wire. In this case, the current is simply found by Ohms Law I = E R 2 = 11 . 2 V 10 = 1 . 12 A . 002 (part 2 of 3) 10.0 points Now the switch is thrown quickly from a to b . Compute the initial voltage across the in- ductor. Correct answer: 1397 . 76 V. Explanation: Let : R 1 = 1238 and I = 1 . 12 A . When the switch is thrown from a to b , the current in the circuit is the current passing through R 2 , which was found in Part 1 to be 1 . 12 A . From Kirchhoffs Loop Law, the initial voltage across the inductor is equal to the initial voltage across R 1 and R 2 . So, we have V L = V R 1 + V R 2 = I R 1 + I R 2 = (1 . 12 A) (1238 ) + (1 . 12 A) (10 ) = 1397 . 76 V . 003 (part 3 of 3) 10.0 points How much time elapses before the voltage across the inductor drops to 12 V? Correct answer: 11 . 4368 ms. Explanation: Let : L = 3 H and V L = 12 V . The voltage across an inductor is V L =- L dI dt . When the switch is at b , we are dealing with an RL circuit with an initial current I that decays as I = I e- t/ . pan (zp695) Ch23-h3 chiu (56565) 2 The time constant is = L R t = L R 1 + R 2 = 3 H 1238 + 10 = 0 . 00240385 s . Therefore, the voltage across the inductor is V L =- L d dt I e- t/ = L I e- t/ . Solving for t , we obtain t =- ln bracketleftbigg V L L I bracketrightbigg =- (0 . 00240385 s) ln bracketleftbigg (12 V) (0 . 00240385 s) (3 H) (1 . 12 A) bracketrightbigg = 0 . 0114368 s = 11 . 4368 ms . 004 10.0 points A(n) 53 . 1 cm length of wire when used as a re- sistor has a resistance of 0 . 00526 . The ends of the wire are connected to form a circular loop, and the plane of the loop is positioned at right angles to a uniform magnetic field that is increasing at the rate of 0 . 0683 T / s....
View Full Document

Page1 / 8

Ch23-h3-solutions - pan (zp695) Ch23-h3 chiu (56565) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online