Soln11 - EE411 Spring 2008 HW#11 Solution (Q2) 11.66 - Zf -...

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EE411 Spring 2008 HW#11 Solution (Q2) 11.66 As an inverter, ) 45 4 ( j3 4 j4) (2 - - s i f o ° + + = = V Z Z V mA j3) j2)(4 - (6 ) 45 j4)(4 (2 - mA 2 j 6 o o + ° + = = V I The power absorbed by the 6-k Ω resistor is 3 6 - 2 2 o 10 6 10 5 40 4 20 2 1 R 2 1 P × × × × × = = I = P mW 96 . 0 (Q4) 12.12 Convert the delta-load to a wye-load and apply per-phase analysis.
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This note was uploaded on 10/26/2009 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas.

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