chapter 11

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

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Chapter 11: AC Steady State Power Exercises: Ex. 11.3-1 4( 2) 3 0.8 1.4 42 1.6 60.3 70 4.38 60.3 A 1.6 60.3 j jj j ° ° ° ° =+ = + =∠ Ω == Z V I = Z ( ) 4.38cos(10 60.3 ) A it 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 pt vt it t t t t =⋅= −° = ° + − ° The inductor voltage is calculated as LL = = (4.38 60.3 )(j3) = 13.12 29.69 V ⋅∠ ° ∠ VI Z ° L ( ) 13.12cos(10 29.69 ) V vt t = The instantaneous power delivered to the inductor is given by ( ) ( ) ( ) (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 pt vti 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 vt V it t RR ωθ == + The instantaneous power delivered to the resistor is given by 2 22 m 2 mm m Rm () () ) ) ) cos(2 ) VV pt v ti t V t t t t R R ω θω θ =⋅= + += +=+ + (b) When the element is an inductor, the current will lag the voltage by 90 ° . L 90 90 90 jL L LL ωω = =∠ ° Ω⇒ = = ∠− ° ∠° V ZI = Z The instantaneous power delivered to the inductor is given by 2 Lm 90) ) cos 2 2 W 2 pt i tv t t V t t ° =⋅ = + −⋅ + ° 11-2
Ex. 11.3-3 The equivalent impedance of the parallel resistor and inductor is ( )( ) () 1 1 1 12 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 22 P θ    == ° = IV (b) 1 2 2 max 1 R 20 1 10 20 W IR P = Ex. 11.4-1 2 3 2 eff 0 02 11 = ( ) (10) (5) 8.66 3 t Ii t d t d t d t T    ∫∫ = Ex. 11.4-2 (a) max eff 2 2cos3 A 2 A I it t I =⇒ = = (b) ( ) cos(3 90 ) 60 ) A t t °° =− ++ ( ) 13 1 90 1 60 0.518 15 A jj −°+∠°= −++ = I eff 0.518 ( ) 0.518 cos (3 15 ) A = 0.366 A 2 t I ° ⇒= (c) 2 eff eff 23 2.55 A II  =   11-3

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Ex. 11.4-3 Use superposition: 1 50 V =∠° V 2 2.5 V (dc) = V 3 39 0 =∠ − ° V V V 1 and V 2 are phasors having the same frequency, so we can add them: () ( ) 13 5 0 39 0 53 5 . 8 33 1 . 0 j V + °+∠ −°=− = ° VV Then R12 3 ( ) ( ) ( ) ( ) 2.5 5.83cos(100 31.0 ) V vt vt vt t =+ + =+ −° Finally eff eff 2 22 RR 5.83 (2.5) 23.24 V 4.82 V 2  =+ = ⇒=   Ex. 11.5-1 Analysis using Mathcad (ex11_5_1.mcd): Enter the parameters of the voltage source: A1 2 := ω 2 := Enter the values of R and L R1 0 := L4 := The impedance seen by the voltage source is: ZR j ω L + := The mesh current is: I A Z := 11-4
The complex power delivered by the source is: Sv I IZ () 2 := Sv 4.39 3.512i + = The complex power delivered to the resistor is: Sr I IR 2 := Sr 4.39 = The complex power delivered to the inductor is: Sl I Ij ⋅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 L 2 := The complex power delivered to the inductor is: Sr 6.606 = Sr I 2 := The complex power delivered to the resistor is: Sv 6.606 1.982i + = Sv I 2 := The complex power delivered by the source is: I A Z := The mesh current is: ZR j ω L + 1 j ω C + := The impedance seen by the voltage source is: C 0.1 := L4 := R1 0 := Enter the values of R, L an d C ω 2 := A1 2 :=

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## This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.

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

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