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Unformatted text preview: gonzales (pag757) Homework 11 Hoffmann (56745) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. After the parts that are withdrawn there are 20 problems. 001 (part 1 of 2) 10.0 points A smalldiameter inductor with self inductance L s and a largediameter inductor with selfinductance L are connected in se ries as shown in the figure. Consider the two inductors to be far apart. Thus, the mutual inductance between the two inductors is 0. a b Determine the equivalent selfinductance for the system. 1. L eq = radicalbig ( L s L ) 2 2 2. L eq = L s L L s + L 3. L eq = radicalBig ( L s L ) 2 4. L eq = L s + L correct 5. L eq = L s L radicalBig L 2 s + L 2 6. L eq = L s L 2 ( L s + L ) 7. L eq = L s L 2 radicalBig L 2 s + L 2 8. L eq = radicalbig ( L s + L ) 2 2 9. L eq = L s + L 2 10. L eq = radicalBig ( L s + L ) 2 Explanation: Let E be the emf across the larger diameter inductor, E s be the emf across the smallerdiameter inductor, and E be the total emf across both inductors. The current I entering terminal a and ex iting terminal b of this series combination is related to the total emf E by E = L eq dI dt , (1) so L eq = E dI dt . (2) In general, the emf E across the inductor complex is due to selfinductance and mutual inductance between the two inductors. How ever, as M = 0 and as I = I s = I , we have E s = L s dI s dt = L s dI dt , (3) and similarly E = L dI dt = L dI dt . (4) Using E = E + E s together with Eqs. 1, 3, and 4, we have L eq dI dt = E + E s = L dI dt L s dI dt = ( L + L s ) dI dt . (5) Therefore for L eq , we have L eq = L + L s . (6) Note: Series inductors combine like series resistors (if no mutual inductance is present). 002 (part 2 of 2) 0.0 points WITHDRAWN 003 (part 1 of 2) 10.0 points gonzales (pag757) Homework 11 Hoffmann (56745) 2 A smalldiameter solenoidal inductor with selfinductance 131 mH and a largediameter solenoidal inductor with selfinductance 54 mH are connected in parallel as shown in the figure. Consider the two inductors to be far apart. Thus, the mutual inductance between the two inductors is 0 mH. a b Determine the equivalent selfinductance for the system. Correct answer: 38 . 2378 mH. Explanation: Let : L s = 131 mH and L = 54 mH . The source of the current through the smaller diameter inductor I s and the largerdiameter inductor I is from the same terminal. That is, the total current I in this parallel inductor complex is the sum of the currents in the two inductors I = I s + I ....
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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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