electrical circuit problem

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

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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 = XC