標題:
electrical circuit problem
發問:
http://i599.photobucket.com/albums/tt72/goldfish_gem/hrthttrhth.jpg
a i) By P = VI, we have I = 300/220 = 1.36 A in phase with the voltage supply ii) From the given, apparent power to the induction motor = 400∠-60 for a power factor of 0.5 lagging. So by P = VI again: I = 400∠-60/220 = 1.82∠-60 A iii) Total current is the sum of current through the bulb and motor, i.e. I (total) = 1.36∠0 + 1.82∠-60 = (5/11)(5 - j2√3) A = 2.76∠-34.7 A iv) Total power = 500W, reactive power (for the motor only) = -200√3 VAR So apparent power = 100(5 - j2√3) W = 608∠-34.7 W b i) Since capacitor only have reactance, it does not draw any resistive component and hence the resistive current of the circuit is still 25/11 = 2.27 A. With a power factor = 0.9, phase angle between voltage and output current = 25.8, i.e. the ratio of: Reactive current/Resistive current = tan 25.8 = 0.484 Reactive current = -j0.484 x 25/11 = -j1.1 (as it is still lagging) Hence new current = (2.27 - j1.1) A = 2.53∠-25.8 A So apparent power = 220 x 2.53∠-25.8 = 555.6∠-25.8 W Total power = 555.6 x 0.9 = 500 W Reactive power = -555.6 sin 25.8 = -242 VAR (lagging) ii) Reactive current drawn by the capacitor = (10√3)/11 - 1.1 = 0.474 A So reactance of the capacitor = -220/0.474 = -j464Ω So we have: 1/(100πC) = 464 C = 6.86 μF
其他解答:
這裏可以幫到你 http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536|||||圖片參考：http://i599.photobucket.com/albums/tt72/goldfish_gem/hrthttrhth.jpg Resistive load = 300W Indutive Load = 200W 0.5Cosφ Voltage = 220V 50Hz B1 (a) i) I = P / V = 300 / 220 = 1.3636A ii) I = P / (VCosφ) = 200 / (220x0.5) = 200 / 110 = 1.8181A iii) IT = √IR^2 + IL^2 = √1.3636^2 + 1.8181^2 = √1.8594 + 3.3055 = √5.1649 = 2.2726A iv) P = 300+200=500W Q = 200*1.732=346.4VAR S = 300+(200/0.5)=700VA B1 (b) i) P = 300+200=500W Q = 200*0.4843=96.9VAR S = 300+(200/0.9)=522.22VA ii) for 0.9Cosφ ==> 1.0Cosφ Q = 200*0.4843=96.9VAR for 0.5Cosφ ==> 0.9Cosφ = 346.4 - 96.9 = 249.5VAR = 250VAR = 0.25kVAR 朋友！到你考試時又點問呀, 應該係老師教你時睡(訓)覺。 2010-05-05 17:42:22 補充： Capacitor Q=V^2/Xc Xc=V^2/Q =(220*220)/250=193.6Ω c=10^6/2ЛFXc =10^6/2*3.142*50*193.6 =10^6/60829.12 =16.44μF so you can use 16.5μF Capacitor 2010-05-05 17:50:28 補充： 只要你记住 (kW) P=V*I*Cosφ (kVAR) Q=P*tanφ (kVA) S=V*I P=V^2/R P=I^2*R Resistive = R Inductive = XL Capacitive = XC63D0B758E2D502CC
electrical circuit problem
發問:
http://i599.photobucket.com/albums/tt72/goldfish_gem/hrthttrhth.jpg
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最佳解答:a i) By P = VI, we have I = 300/220 = 1.36 A in phase with the voltage supply ii) From the given, apparent power to the induction motor = 400∠-60 for a power factor of 0.5 lagging. So by P = VI again: I = 400∠-60/220 = 1.82∠-60 A iii) Total current is the sum of current through the bulb and motor, i.e. I (total) = 1.36∠0 + 1.82∠-60 = (5/11)(5 - j2√3) A = 2.76∠-34.7 A iv) Total power = 500W, reactive power (for the motor only) = -200√3 VAR So apparent power = 100(5 - j2√3) W = 608∠-34.7 W b i) Since capacitor only have reactance, it does not draw any resistive component and hence the resistive current of the circuit is still 25/11 = 2.27 A. With a power factor = 0.9, phase angle between voltage and output current = 25.8, i.e. the ratio of: Reactive current/Resistive current = tan 25.8 = 0.484 Reactive current = -j0.484 x 25/11 = -j1.1 (as it is still lagging) Hence new current = (2.27 - j1.1) A = 2.53∠-25.8 A So apparent power = 220 x 2.53∠-25.8 = 555.6∠-25.8 W Total power = 555.6 x 0.9 = 500 W Reactive power = -555.6 sin 25.8 = -242 VAR (lagging) ii) Reactive current drawn by the capacitor = (10√3)/11 - 1.1 = 0.474 A So reactance of the capacitor = -220/0.474 = -j464Ω So we have: 1/(100πC) = 464 C = 6.86 μF
其他解答:
這裏可以幫到你 http://actioni.w7.c361.net/yahoo.com.hk/hk/auction/178987536|||||圖片參考：http://i599.photobucket.com/albums/tt72/goldfish_gem/hrthttrhth.jpg Resistive load = 300W Indutive Load = 200W 0.5Cosφ Voltage = 220V 50Hz B1 (a) i) I = P / V = 300 / 220 = 1.3636A ii) I = P / (VCosφ) = 200 / (220x0.5) = 200 / 110 = 1.8181A iii) IT = √IR^2 + IL^2 = √1.3636^2 + 1.8181^2 = √1.8594 + 3.3055 = √5.1649 = 2.2726A iv) P = 300+200=500W Q = 200*1.732=346.4VAR S = 300+(200/0.5)=700VA B1 (b) i) P = 300+200=500W Q = 200*0.4843=96.9VAR S = 300+(200/0.9)=522.22VA ii) for 0.9Cosφ ==> 1.0Cosφ Q = 200*0.4843=96.9VAR for 0.5Cosφ ==> 0.9Cosφ = 346.4 - 96.9 = 249.5VAR = 250VAR = 0.25kVAR 朋友！到你考試時又點問呀, 應該係老師教你時睡(訓)覺。 2010-05-05 17:42:22 補充： Capacitor Q=V^2/Xc Xc=V^2/Q =(220*220)/250=193.6Ω c=10^6/2ЛFXc =10^6/2*3.142*50*193.6 =10^6/60829.12 =16.44μF so you can use 16.5μF Capacitor 2010-05-05 17:50:28 補充： 只要你记住 (kW) P=V*I*Cosφ (kVAR) Q=P*tanφ (kVA) S=V*I P=V^2/R P=I^2*R Resistive = R Inductive = XL Capacitive = XC63D0B758E2D502CC
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