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chapter 11

# chapter 11 - Chapter 11 AC Steady State Power Exercises Ex...

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Chapter 11: AC Steady State Power Exercises: Ex. 11.3-1 4( 2) 3 0.8 1.4 4 2 1.6 60.3 7 0 4.38 60.3 A 1.6 60.3 j j j j ° ° ° ° = + = + = = = ∠ − Z V I = Z ( ) 4.38cos(10 60.3 ) A i t t ° = The instantaneous power delivered by the source is given by [ ] (7)(4.38) ( ) ( ) ( ) (7cos10 )(4.38cos(10 60.3 )) cos(60.3 ) cos(20 60.3 ) 2 7.6 15.3cos(20 60.3 ) W p t v t i t t t t t = = ° = ° + ° = + ° The inductor voltage is calculated as L L = = (4.38 60.3 )(j3) = 13.12 29.69 V ∠− ° V I Z ° L ( ) 13.12cos(10 29.69 ) V v t t = + ° The instantaneous power delivered to the inductor is given by [ ] L L ( ) ( ) ( ) (13.12 cos (10 29.69 )(4.38cos(l0t 60.3 ) 57.47 cos(29.69 60.3 ) cos(20 29.69 60.3 ) 2 28.7 cos (20 30.6 ) W p t v t i t t t t ° = = + ° = °+ ° + + ° = ° ° 11-1

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Ex. 11.3-2 (a) When the element is a resistor, the current has the same phase angle as the voltage: m ( ) ( ) cos( ) A v t V i t t R R ω θ = = + The instantaneous power delivered to the resistor is given by 2 2 2 m 2 m m m R m ( ) ( ) ( ) cos( ) cos( ) cos ( ) cos(2 ) 2 2 V V V V p t v t i t V t t t t R R R R ω θ ω θ ω θ ω = = + + = + = + + θ (b) When the element is an inductor, the current will lag the voltage by 90 ° . ( ) m m L 90 90 90 V V j L L L L θ ω ω θ ω ω = = ° Ω = = ° ° V Z I = Z The instantaneous power delivered to the inductor is given by ( ) 2 m m L m ( ) ( ) ( ) cos( 90 ) cos( ) cos 2 2 90 W 2 V V p t i t v t t V t t L L ω θ ω θ ω θ ω ω ° = = + + = + ° 11-2
Ex. 11.3-3 The equivalent impedance of the parallel resistor and inductor is ( )( ) ( ) 1 1 1 1 2 j j j = = + + Z . Then ( ) 10 0 20 18.4 A 1 10 1+ 1 2 j ° = = ∠ − ° + I (a) ( ) ( ) source 20 10 10 cos cos 18.4 30.0 W 2 2 P θ = = ° = I V (b) ( ) 1 2 2 max 1 R 20 1 10 20 W 2 2 I R P = = = Ex. 11.4-1 2 3 2 2 2 eff 0 0 2 1 1 = ( ) (10) (5) 8.66 3 t I i t dt dt dt T = + = Ex. 11.4-2 (a) max eff 2 ( ) 2cos3 A 2 A 2 2 I i t t I = = = (b) ( ) cos(3 90 ) cos(3 60 ) A i t t t ° ° = + + ( ) ( ) 1 3 1 90 1 60 0.518 15 A 2 2 j j = ∠− ° + ° =− + + = ∠ − ° I eff 0.518 ( ) 0.518 cos (3 15 ) A = 0.366 A 2 i t t I = ° = (c) 2 2 2 eff eff 2 3 2.55 A 2 2 I I = + = 11-3

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Ex. 11.4-3 Use superposition: 1 5 0 V = ∠ ° V 2 2.5 V (dc) = V 3 3 90 = ∠− ° V V V 1 and V 2 are phasors having the same frequency, so we can add them: ( ) ( ) 1 3 5 0 3 90 5 3 5.83 31.0 j V + = ∠ ° + ∠− ° = = ∠− ° V V Then ( ) R 1 2 3 ( ) ( ) ( ) ( ) 2.5 5.83cos(100 31.0 ) V v t v t v t v t t = + + = + ° Finally eff eff 2 2 2 R R 5.83 (2.5) 23.24 V 4.82 V 2 V V = + = = Ex. 11.5-1 Analysis using Mathcad (ex11_5_1.mcd): Enter the parameters of the voltage source: A 12 := ω 2 := Enter the values of R and L R 10 := L 4 := The impedance seen by the voltage source is: Z R j ω L + := The mesh current is: I A Z := 11-4
The complex power delivered by the source is: Sv I I Z ( ) 2 := Sv 4.39 3.512i + = The complex power delivered to the resistor is: Sr I I R ( ) 2 := Sr 4.39 = The complex power delivered to the inductor is: Sl I I j ⋅ ω L ( ) 2 := Sl 3.512i = Verify Sv = Sr + Sl : Sr Sl + 4.39 3.512i + = Sv 4.39 3.512i + = Ex. 11.5-2 Analysis using Mathcad (ex11_5_2.mcd): Sv 6.606 1.982i + = Sr Sl + Sc + 6.606 1.982i + = Verify Sv = Sr + Sl + Sc : Sc 3.303i = Sc I I 1 j ω C 2 := The complex power delivered to the capacitor is: Sl 5.284i = Sl I I j ⋅ ω L ( ) 2 :=

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chapter 11 - Chapter 11 AC Steady State Power Exercises Ex...

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