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3 Pages

### hw3soln

Course: EECE 253, Spring 2011
School: UBC
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Word Count: 544

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all Note solutions assume rms values unless otherwise specified. 1) Transform the circuit: 4 I I1 2 I2 2j -2j Vp = 10 30V For the 4 resistor: 1 2 1 P = I R = (1.581) 2 ( 4) = 5 W 2 2 For the 2 resistor: 1 2 1 2 P = I 2 R = (1.581) ( 2) = 2.5 W 2 2 For the inductor and the capacitor: The total impedance seen by the voltage source is: Z = 4 + ( j2) || (2 - j2) ( j2)(2 - j2) = 4 + 2 + j2 j2 + 2 - j2 = 6 + j2 =...

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all Note solutions assume rms values unless otherwise specified. 1) Transform the circuit: 4 I I1 2 I2 2j -2j Vp = 10 30V For the 4 resistor: 1 2 1 P = I R = (1.581) 2 ( 4) = 5 W 2 2 For the 2 resistor: 1 2 1 2 P = I 2 R = (1.581) ( 2) = 2.5 W 2 2 For the inductor and the capacitor: The total impedance seen by the voltage source is: Z = 4 + ( j2) || (2 - j2) ( j2)(2 - j2) = 4 + 2 + j2 j2 + 2 - j2 = 6 + j2 = 4 + P = 0W Using this we can calculate I: V 10 30 I = = = 1.58111.565 A 6 + j2 Z Now we use current dividers to calculate I1 and I2: j2 I1 = I = 1.581101.565 A 2 2 - j2 I2 = I = 2.236 - 33.435 A 2 For the source: S = 1 * 1 V I = (1030)(1.581 - 11.565) 2 2 = 7.5 + j 2.5 VA P = 7.5 W j1 S 2) 3) 0.5 S V1 4 60 A + V0 - V2 + 4V0 - Mathematically the function can be written as: V3 A -0.5j S 0.25 S t 0 < t < 2 v(t) = - 1 2 < t < 4 RMS voltage is calculated by: Convert the circuit and form the matrix for modified nodal analysis: -0.5 -j 0.5 + j V1 -0.5 V = 0.5 - j 0.5 0 2 -j 0 0.25 + j V3 Using conditions on dependent voltage source: 4V0 = V2 - V3 V0 = V1 - V2 4V1 - 5V2 + V3 = 0 4 60 A -A 2 Vrms = 4 1 2 2 2 t dt + ( -1) dt = 1.1667 V 4 0 2 Vrms = 1.08 V Expand the matrix and add the new condition: 0 V1 -0.5 -j 0.5 + j 4 60 0.5 - j 0.5 0 0 - 1 V2 -0.5 = -j 0 0 0.25 + j 1 V3 1 0 A -5 4 0 Solving for V3: V3 = 9.992 5.06 V For the 4 resistor: P = V32 R = 24.96 W 1 4) 5) Using the figure given in the textbook & nodal analysis: At node V1: 120 30 - V1 V1 V - V2 = + 1 20 j30 50 At node V2: V1 - V2 V V2 = 2 + 50 10 -j40 Using these eqns we can find: V1 = 45.04 + j 66.935, (a): Pj 30 = P- j 40 = 0 P = 10 P50 = P20 = V22 = 17.33 W R a) The total input impedance is: Z = 4 + ( (- j2) || (j5 - j2) ) ( - j2)( j3) = 4 + = 4 - j6 - j2 + j3 = 7.211 -56.31 Which a gives power factor of: pf = cos ( -56.31) = 0.5547 leading V 2 = 9.423 + j 9.913 b) The total input impedance is: Z = 1 || ( - j1 + ( ( j2 ) || (4 + j1 ) ) ) ( - j1 + 0.64 + j1.52 |V 1 |120 30 - V | 1 - V2 | = 92.06 W R 2 2 ( j2)( 4 + j1) = 1 || = 1 || - j1 + 5 + j3 (1)( 0.64 + j0.52 ) = 0.4793 21.5 = 1.64 + j0.52 ) Which gives a power factor of: pf = cos ( 21.5) = 0.9304 lagging R = 175.72 W (b): 120 30 - V1 20 S = VS I * = 285.0 + j 212.6, I = pf = 285 = 0.8015 lagging 355.6 S = 355.6 VA (c): 6) 7) Redraw circuit: transform voltage source into its Norton equivalent, and transform impedances into admittances V1 I1 I2 a) The total input impedance is: Z = 2 + (10 - j5) || (8 + j6) (10 - j5)(8 + j6) 18 + j1 = 8.188 5.382 = 2 + 4.8 0 A 0.02S j0.05S 40A 0.016 - j 0.012 S Which gives a power factor of: pf = cos ( 5.382 ) = 0.9956 lagging b) Average power is: 2 * V2 (16 ) S = V I = = * ( 8.188 -5.382 ) Z = 31.26 5.382 VA = 31.12 + j 2.93 VA Form node equation: [ 0.02 V1 + j 0.05 + 0.016 - j 0.012] V1 = [ 4.8 + 4.0] 8.8 = 0.036 + j 0.038 V1 = 168.13 -46.5 V Now find the currents across the inductor and capacitor: I1 = V1 ( j 0.05) = 8.4143.45 A P = Re S ( ) = 31.12 W b) Reactive power is: Q = Im S ( ) = 2.93 VAR c) Apparent power is: S = 31.26 VA d) Complex power is: S = 31.26 5.382 VA I 2 = V1 (0.016 - j 0.012) = 3.36 - 83.42 A Reactive power in the inductor is: S = I2Z = ( 3.363) ( j30 ) 2 = j339.3 VAR Reactive power in the capacitor is: S = I 2 Z = (8.41) (- j 20 ) = - j1414 VAR 2 2 8) I2 I1 8 120 0 V j12 IS Using nodal analysis: IS = I1 - I2 I1 = 120 0 = 4.62 - j6.92 A 8 + j12 Now using the complex power: S = VI2 * 2500 - j400 I2* = = 20.83 - j3.33 120 0 I2 = 20.83 + j3.33 A IS = I1 - I2 = ( 4.62 - j6.92) - ( 20.83 + j3.33) = -16.21 - j19.54 A = -16.21 - j10.26 A 3
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UBC - EECE - 253
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