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Unformatted text preview: Homework 11 Solutions P11.67 (a) ( 29 1 VI=220 7.6 1672 VA P 1317 pf .788 VI 1672 =cos pf 38.0 Q=VIsin =1030VAR θ θ ° = = = = = ⇒ (b) To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing – Q, then 2 2 c V (220) 1030 = = X = 47 1 1 C = = = 56.5 F X (377)(47) c c X X μ ϖ ⇒ Ω ∴ (c) * = VI cos where = 0 then 1317 = 220I I = 6.0A for corrected pf Note I = 7.6A for uncorrected pf P θ θ ° ∴ P11.68 First load: 1 1 (1 tan (cos (.6))) 500(1 tan53.1 ) 500 677 kVA P jQ P j j j = + = + = + ° = + S Second load: 2 400 600 kVA j = + S Total: 1 2 900 1277 kVA j = + S = S +S 1 desired tan (cos (.90)) 900 436 VA P jP j = + = + S desired From the vector diagram: Q = + S S . Therefore 900 436 900 1277 841 VAR j j Q Q j + = + + ⇒ =  2 2 2 * * (1000) j841 1189 1189 j841 841 377 j j j j C =  ⇒ = = = ⇒ =  =  V V Z Z Z Finally, 1 2.20 F (1189)(377) C μ = = P11.81 t * L t 4000  2000 800 1600 800 1600 800 1000 800 1600 1.6 H j j j R R j L j L =...
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This note was uploaded on 09/30/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Tech.
 Spring '08
 JUANG

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