EE2111_A03_SOL7

EE2111_A03_SOL7 - EE2111 Homework#7 A03 Assigned 09 October...

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EE2111 Homework #7 A03 Assigned 09 October 03 Due 15 October 03 Problem 1 . Hambly 15.29 Solution: () mH N L 80 10 5 200 5 2 2 = × = = Problem 2. Hambly 15.30 Solution: The impedance of an inductance to a sinusoidal current is H j j j Z L j I V Z L L L L 3183 . 0 60 2 90 90 90 1 120 0 = × = = = = = π ω The reluctance is found by rearranging equation 15.25 () 4 2 2 10 54 . 78 3183 . 0 500 × = = = L N Rearranging equation 15.21, we have 3 . 405 10 4 10 3 . 509 10 3 . 509 10 5 10 54 . 78 10 20 7 6 0 6 4 4 2 = × × = = × = × × × × = = µ r A l

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Problem 3. Hambly 15.35 Solution: Because coil 2 is open circuited, ( ) 0 2 = t i . Thus we have () 12 1 1 = dt t di L Integrating and using the fact that ( ) 0 0 1 = i , we have () ( ) t t L t i 120 / 12 1 1 = = Substituting into equation 15.37, we have () () V dt t di M t v 120 1 2 = = Problem 4. Hambly 15.45 Solution: If residential power was distributed at 12 V rather than 120 V higher currents (by an order of magnitude) would be required to deliver the same amounts of power. This would require
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This note was uploaded on 10/13/2009 for the course ECE 2111 taught by Professor Labonte during the Spring '09 term at WPI.

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EE2111_A03_SOL7 - EE2111 Homework#7 A03 Assigned 09 October...

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