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# soltfmr - ECE225 LINEAR TRANSFORMER SOLUTIONS jX M=j8 j5.6...

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ECE225 LINEAR TRANSFORMER – SOLUTIONS SPRING 2008 + M jX =j8 2 jX =j5 2 M j(X -X )=-j3 1 M M jX =j8 j(X -X )=j12 1 jX =j20 - T jX =j2.5 R =2.5 ~ T o V =80 0 VAC VAC VAC 256 0 256 0 L - + o j5.6 - + o L R L R j5.6 (a) Find the Thevenin voltage ˜ V T . ˜ V T = j 8 j 5 . 6 + j 12 + j 8 × 256 6 0 o = 8 25 . 6 × 256 6 0 o = 80 6 0 o V (b) Find the Thevenin impedance ¯ Z T = jX T . ¯ Z T = - j 3 + ( j 5 . 6 + j 12)( j 8) j 5 . 6 + j 12 + j 8 = - j 3 + j 5 . 5 = 0 + j 2 . 5 Ω = R T + jX T (c) Find the maximum real power P L,max dissipated in the optimal resistor R L = | X T | . P L,max = | ˜ V T | 2 2 R T + q R 2 T + X 2 T · = | ˜ V T | 2 2 | X T | = 80 2 2 × 2 . 5 = 1280 W 1

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ECE225 IDEAL TRANSFORMER – SOLUTIONS SPRING 2008 + - j5.6 R L o + - j5.6 VAC R L + - 4R = L 10 V =160 0 VAC T ~ o jX =j10 T VAC o j16 j16 21 21 j4 j4 j4 j1 256 0 256 0 (a) Find the Thevenin voltage
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soltfmr - ECE225 LINEAR TRANSFORMER SOLUTIONS jX M=j8 j5.6...

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