hw17sol - Purdue University, School of Electrical and...

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #17 Solution (65–68) Problem 65 (a) v S (t) L 1 L 2 R s R A B k L 1 = 0 . 8 H, L 2 = 0 . 45 H, M = 0 . 175 H, R s = R = 12 Ω , v s ( t ) = 30cos ( 10 t ) V The maximum instantaneous steady state power delivered to R is P R , max = ( max ( i steadystate ) ) 2 R . Since we are interested in the steady state response, we can use phasor method which is usually simpler than the full laplace method. To find the transfer function I ( s ) V s ( s ) , we can write a loop equation in s-domain. - V s ( s ) + R s I + ( L 1 sI MsI ) + RI + ( L 2 sI MsI ) = 0 upper and lower signs correspond to dot in position A and B respectively. I = V s ( s ) ( R s + R ) + ( L 1 + L 2 2 M ) s H ( s ) = I ( s ) V s ( s ) = 1 ( L 1 + L 2 2 M ) s + ( R s + R ) (65.1) (i) Dot in position A Substituting component values into eq.(65.1) for dot in position A, H A ( s ) = 1 0 . 9 s + 24 H A ( j 10 ) = 1 24 + j 9 = 0 . 039 - 20 . 556 i steadystate ( t ) = 30 | H A ( j 10 ) | cos ± 10 t + ( H A ( j 10 )) ² = 1 . 1704cos ( 10 t - 20 . 556 ) [A] i max = 1 . 1704 P R , max = i 2 max R = ( 1 . 1704 ) 2 ( 12 ) = 16 . 438[W] 1/ 13
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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (ii) Dot in position B Following the same steps as in the previous part, H B ( s ) = 1 1 . 6 s + 24 H B ( j 10 ) = 1 24 + j 16 = 0 . 0347 - 33 . 69 i steadystate ( t ) = 30 | H B ( j 10 ) | cos ± 10 t + ( H B ( j 10 )) ² = 1 . 0401cos ( 10 t - 33 . 6901 ) [A] i max = 1 . 04 P R , max = i 2 max R = ( 1 . 04 ) 2 ( 12 ) = 12 . 98[W] (b) Equivalent s-domain circuit accounting for the initial condition: V S (s) R Ls 1/Cs Ls v C (0 )/s k + V C (s) _ I L (s) L = 1 H, R = 3 Ω , C = 1 / 2 F, M = k LL = 0 . 5 H, V s ( s ) = 1 Writing a loop equation, - V s + RI L + ( LsI L - MsI L ) + v C ( 0 - ) s + 1 Cs I L + ( LsI L - MsI L ) = 0 I L ( s ) = V s - v C ( 0 - ) s ( 2 L - 2 M ) s + R + 1 Cs = s 2 ( L - M ) s 2 + R 2 ( L - M ) s + 1 2 ( L - M ) C ³ V s - v C ( 0 - ) s ´ I L ( s ) = s s 2 + 3 s + 2 s - 2 s = 4 s + 2 - 3 s + 1 i L ( t ) = ( 4 e - 2 t - 3 e - t ) u ( t ) [A] 2/ 13
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hw17sol - Purdue University, School of Electrical and...

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