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Unformatted text preview: gonzales (pag757) Homework 11 Hoffmann (56745) 1 This print-out should have 23 questions. Multiple-choice 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 After the parts that are withdrawn there are 20 problems. 001 (part 1 of 2) 10.0 points A small-diameter inductor with self- inductance L s and a large-diameter inductor with self-inductance L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 self-inductance 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 smaller-diameter 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 self-inductance 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Note: Series inductors combine like series resistors (if no mutual inductance is present). 002 (part 2 of 2) 0.0 points WITHDRAWN 003 (part 1 of 2) 10.0 points gonzales (pag757) Homework 11 Hoffmann (56745) 2 A small-diameter solenoidal inductor with self-inductance 131 mH and a large-diameter solenoidal inductor with self-inductance 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 self-inductance 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 larger-diameter 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