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Homework 11-solutions

# Homework 11-solutions - gonzales(pag757 Homework 11...

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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 After the parts that are withdrawn there are 20 problems. 001 (part 1 of 2) 10.0 points A small-diameter inductor with self- inductance L s and a large-diameter inductor with self-inductance L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 d I dt , (1) so L eq = - E d I 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 d I s dt = - L s d I dt , (3) and similarly E = - L d I dt = - L d I dt . (4) Using E = E + E s together with Eqs. 1, 3, and 4, we have - L eq d I dt = E + E s = - L d I dt - L s d I dt = - ( L + L s ) d I 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Note: Series inductors combine like series resistors (if no mutual inductance is present). 002 (part 2 of 2) 0.0 points WITHDRAWN 003 (part 1 of 2) 10.0 points

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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 . The emf E is the same across both induc- tors because they are connected in parallel. The emf is related to the total current I by E = - L eq d I dt . (1) Therefore L eq = - E d I dt . (2) In general the total induced emf E across an inductor is due to self-inductance with itself and mutual-inductance with the other inductor. But with M = 0 mH, we only have self-inductance, so E = - L s d I s dt (3) d I s dt = - E L s . (4) Similarly E = - L d I dt (5) d I dt = - E L . (6) Using Eqs. 4 and 6 to solve for d I dt , where I = I s + I , we have d I dt = d I
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