对称三相负载联结成三角形,接对称三相电源,已知电源线电压为380V,每相负载阻抗R=150Ω,X=200Ω.求线电流

2024-11-27 23:07:21
推荐回答(1个)
回答1:

解:t=0-时,电容相当于开路。Us1——R1——R2——Us2构成回路,回路电流为:
I=(Us1-Us2)/(R1+R2)=(10-5)/(3+2)=1(mA)。顺时针方向。
所以:Uc(0-)=IR2+Us2=1×2+5=7(V)。
换路定理:Uc(0+)=Uc(0-)=7V。
t=∞时,电容再次相当于开路,R2中无电流、无电压,因此Uc(∞)=Us2=5V。
Us2短路,从C看进去的等效电阻:R=R2=2kΩ,所以电路的时间常数为:τ=RC=2000×5/1000000=0.01(s)。
三要素法:Uc(t)=Uc(∞)+[Uc(0+)-Uc(∞)]e^(-t/τ)=5+(7-5)e^(-t/0.01)=5+2e^(-100t) (V)。

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