8 R eq Z cos \u03b8 201 k \u03a9 cos 238 184 k \u03a9 X C eq Z sin \u03b8 201 k \u03a9 sin 238 813 k \u03a9 C

8 r eq z cos θ 201 k ω cos 238 184 k ω x c eq z

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= 23.8 R eq = Z cos θ = (20.1 k Ω )cos 23.8 ° = 18.4 k Ω X C (eq) = Z sin θ = (20.1 k Ω )sin 23.8 ° = 8.13 k Ω C eq = (eq) 2 1 C fX π = 196 pF 23. X C 1 = ) F 1 . 0 )( kHz 15 ( 2 1 μ π = 106 Ω X C 2 = ) F 047 . 0 )( kHz 15 ( 2 1 μ π = 226 Ω X C 3 = ) F 22 . 0 )( kHz 15 ( 2 1 μ π = 48.2 Ω The total resistance in the resistive branch is R tot = R 1 + R 2 = 330 Ω + 180 Ω = 510 Ω The combined parallel capacitance of C 2 and C 3 is C ( tot ) p = C 1 + C 2 = 0.047 μ F + 0.22 μ F = 0.267 μ F X C ( tot ) p = ) F 267 . 0 )( kHz 15 ( 2 1 μ π = 39.7 Ω The impedance of R tot in parallel with C (tot)p is Z p = 2 2 2 ) ( 2 ) ( ) 7 . 39 ( ) 510 ( ) 7 . 39 )( 510 ( Ω + Ω Ω Ω = + p tot C tot p tot C tot X R X R = 39.6 Ω The angle associated with R tot and C (tot)p in parallel is θ = tan 1 Ω Ω = 7 . 39 510 tan 1 ) ( p tot C tot X R = 85.5 ° Converting from parallel to series: R eq = Z p cos θ = (39.6 Ω )cos 85.5 ° = 3.08 Ω X C (eq) = Z p sin θ = (39.6 Ω )sin 85.5 ° = 39.5 Ω The total circuit impedance is Z tot = 2 2 2 ) eq ( 1 2 eq ) 6 . 145 ( ) 08 . 3 ( ) ( Ω + Ω = + + C C X X R = 145.6 Ω SECTION 10-6 Analysis of Series-Parallel RC Circuits Full file at
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100 The voltage across the parallel branches is V C 2 = V C 3 = V R 1 R 2 = V 12 6 . 145 6 . 39 Ω Ω = s tot p V Z Z = 3.26 V V R 1 = V 3.26 510 330 Ω Ω = 2.11 V V R 2 = V 3.26 510 180 Ω Ω = 1.15 V V C 1 = 1 106 12 V 12 V 145.6 C tot X Z Ω = Ω = 8.74 V 24. From Problem 23, R eq = 3.08 Ω and X C 1 + X C (eq) = 145.6 Ω . Since 145.6 Ω > 3.08 Ω , the circuit is predominantly capacitive. 25. Using data from Problem 23: I tot = Ω = 6 . 145 V 12 tot s Z V = 82.4 mA I C 2 = Ω = 226 V 3.26 2 2 C C X V = 14.4 mA I C 3 = Ω = 2 . 48 V 3.26 3 3 C C X V = 67.6 mA I R 1 = I R 2 = Ω 510 V 3.26 = 6.39 mA 26. R tot = R 1 + R 2 || R 3 = 89.9 Ω X C = ) F 47 . 0 )( kHz 1 ( 2 1 μ π = 339 Ω Z tot = 2 2 2 2 (89.9 ) (339 ) tot C R X + = Ω + Ω = 351 Ω (a) I tot = Ω = 351 V 15 tot s Z V = 42.7 mA (b) θ = tan 1 1 339 tan 89.9 C tot X R Ω = Ω = 75.1 ° ( I tot leads V s ) (c) V R 1 = V 15 351 47 1 Ω Ω = s tot V Z R = 2.01 V (d) V R 2 = V 15 351 9 . 42 3 2 Ω Ω = s tot V Z R R = 1.83 V (e) V R 3 = V R 2 = 1.83 V (f) V C = V 15 351 339 Ω Ω = s tot C V Z X = 14.5 V Full file at
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101 27. P a = 2 2 2 2 true ) VAR 3.5 ( ) W 2 ( + = + r P P = 4.03 VA 28. From Problem 10: I tot = 103 mA, X C = 79.6 Ω P rue = R I tot 2 = (103 mA) 2 (56 Ω ) = 0.591 W P r = C tot X I 2 = (103 mA) 2 (79.6 Ω ) = 0.840 VAR 29. Using the results from Problem 22: R eq = 18.4 k Ω X C (eq) = ) F p 196 )( kHz 100 ( 2 1 2 1 eq π = π fC = 8.13 k Ω θ = tan 1 Ω Ω = k 4 . 18 k 13 . 8 tan 1 eq ) eq ( R X C = 23.8 ° PF = cos θ = cos 23.8 ° = 0.915 30. From Problem 26: I tot = 42.7 mA, R tot = 89.9 Ω , X C = 339 Ω , Z tot = 351 Ω P true = tot tot R I 2 = (42.7 mA) 2 (89.9 Ω ) = 169 mW P r = C tot X I 2 = (42.7 mA) 2 (339 Ω ) = 6 18 mVAR P a = tot tot Z I 2 = (42.7 mA) 2 (351 Ω ) = 640 mVA PF = cos = cos(75.1 ) = 0.257 31. Use the formula, V out = V. 1 tot C Z X See Figure 10-2.
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