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ACCA08short

ACCA08short - ACCA 1 AC CIRCUIT ANALYSIS AC SERIES CIRCUIT...

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ACCA AC CIRCUIT ANALYSIS AC SERIES CIRCUIT ANALYSIS Impedance ( Z ~ ) analogous to total resistance of a circuit ~ ~ ~ Z V I = ~ V S = 40 20 ° ~ ~ ~ V R I I R = = 6 ~ ~ ( )( ~ ~ V j L I j I j I L = = = ϖ 4 2) 8 ~ ~ ( . ) ~ ~ V j C I j I j I C = - = - = - 1 1 4 0 0625 4 ϖ ~ ~ ~ ~ V V V V S R L C = + + KVL 40 20 ° = ( ) ~ 6 8 4 + - j j I = ( ) ~ 6 4 + j I = 7.21 33.7 ° I ~ I ~ = 40 20 ° 7.21 33.7 ° = 5.548 -13.7 ° I ~ = 5 548 2 . sin (4t - 13.7 ° ) A 1 6 2 H 62.5 mF V S = sin (4t + 20 ° ) ) 2 ( ) 4 ( j L j L Z ~ = ϖ = ) 2 ( ) 4 ( j L j L Z ~ = ϖ = ) 2 ( ) 4 ( j L j L Z ~ = ϖ = i

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ACCA Alternate Solution ~ Z R R = = 6 ) 2 ( ) 4 ( j L j L Z ~ = ϖ = Z j C j j C ~ ( . ) = - = - = - ϖ 4 0 0625 4 Z Z Z Z T R L C ~ ~ ~ ~ = + + = 6 + j 8 - j 4 = 6 4 2 2 + tan -1 ( 4 / 6 ) = 7.21 33.7 ° I V Z S T ~ ~ ~ = = 40 20 ° 7.21 33.7 ° = 5.548 -13.7 ° A AC PARALLEL CIRCUIT ANALYSIS 2
ACCA I V R R ~ = = 10 0 ° 1000 = 0.01 0 ° A I L ~ = 10 0 ° j (5000)(0.5) = 10 0 ° 2500 90 ° = 0.004 -90 ° A I C ~ = 10 0 ° 1 ϖ C -90 ° = 10 0 ° 1000 -90 ° = 0.01 90 ° A I I I I S R L C ~ ~ ~ ~ = + + = 0.01 - j 0.004 + j 0.01 = 0.01 + j 0.006 A = 0.0117 31 ° A 0 01 0 006 0 0117 2 2 . . . + = tan . . . θ θ = = = 0 006 0 01 0 6 31.0 ° Alternate Solution ~ Z R R = = 1000 ~ ( )( . ) Z j L j j L = = = ϖ 5000 0 5 2500 3 1 k 0.2 μ F V S =sin (5000t) ) 2 ( ) 4 ( j L j L Z ~ = ϖ = ) 2 ( ) 4 ( j L j L Z ~ = ϖ = ) 2 ( ) 4 ( j

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ACCA08short - ACCA 1 AC CIRCUIT ANALYSIS AC SERIES CIRCUIT...

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