pra #5

# pra #5 - ECSE-210 Winter 2011 Electric Circuits 1 Practice...

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ECSE-210 Winter 2011 Electric Circuits 1 5-1 Practice Set #5 P11.6-7 (Using all rms values) (a) ( ) 1 220 7.6 1672 VA 1317 .788 1672 =cos pf 38.0 sin =1030 VAR VI P pf VI Q VI ! " ° = = = = = = # = (b) To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing – Q , then 2 2 c (220) 1030 = = X = 47 1 1 = = = 56.5 F (377)(47) c c V X X C X μ " # \$ (c) * cos where = 0 then 1317 = 220I = 6.0Arms for corrected Note 7.6Arms for uncorrected . P VI I pf I pf ° = " = P11.6-8 First load: ( ) ( ) ( ) ( ) 1 1 1 tan cos .6 500 1 tan 53.1 500 667 kVA P jQ P j j j ! = + = + = + ° = + S Second load: 2 400 600 kVA j = + S Total: 1 2 900 1267 kVA j = + S = S +S 1 desired tan (cos (.90)) 900 436 VA P jP j ! = + = + S desired From the vector diagram: jQ = + S S . Therefore 900 436 900 1277 841 VAR j j jQ Q + = + + ! = " 2 2 2 * * (1000) 841 1189 1189 841 841 377 j j j j j j C = ! " = = = " = ! = ! ! ! V V Z Z Z Finally, 1 2.20 F (1189)(377) C = =

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ECSE-210 Winter 2011 Electric Circuits 1 5-2 P11.6-13 Let 2 2 1 tan X R j X R X R ! " # = + = + \$ % ' ( Z be the impedance of the load. Further, let A ! = " V and B = " I be the voltage across and current through the load. Then ( ) 2 2 1 tan X A A R X R j X R B B " # \$ % + \$ = + = = = \$ # ' ( \$ ) * Z Equating angles gives 1 tan X R # \$ % = # ' ( ) Also, the complex power delivered to the load is
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