三相四线制电路典型分析

三相四线制电路典型分析

如图所示的三相四线制电路，电源线电压 $U_L=380V$；三个电阻性负载连成星形，其电阻为 $R_a=11\Omega ,R_b=R_c=22\Omega$。试求：
（1）负载电压，相电流及中性线电流
（2）如无中性线，求中性点 N’ 电压及负载电压
（3）如无中性线，当 A 相短路时求各相电压和电流
（4）如无中性线，当 C 相断路时求另外两相的电压和电流
（5）在（3）、（4）中如有中性线，则又如何

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答：

（1）负载电压、相电流及中性线电流

$$U_P=\frac{U_L}{\sqrt{3}}=\frac{380}{\sqrt{3}}=220V$$
$$\dot{U_A}=220\angle 0^{\circ }V$$
$$\dot{U_B}=220\angle (-120^{\circ })V$$
$$\dot{U_C}=220\angle 120^{\circ }V$$
$$\dot{I_A}=\frac{\dot{U_A}}{R_a}=\frac{220\angle 0^{\circ }}{11}=20\angle 0^{\circ }A$$
$$\dot{I_B}=\frac{\dot{U_B}}{R_b}=\frac{220\angle (-120^{\circ })}{22}=10\angle (-120^{\circ })A$$
$$\dot{I_C}=\frac{\dot{U_C}}{R_c}=\frac{220\angle 120^{\circ }}{22}=10\angle 120^{\circ }A$$

$$\dot{I_N}=\dot{I_A}+\dot{I_B}+\dot{I_C}=20\angle 0^{\circ }+10\angle (-120^{\circ })+10\angle 120^{\circ }=10\angle 0^{\circ }A$$

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（2）无中性线时，中性点 N’ 电压及负载电压

设 N’ 点电压 $\dot{U_{N^{\prime }}}=x\angle \theta V$，则 $\dot{U_{N^{\prime }}}=x\cos \theta +jx\sin \theta$ V

$$\begin{array}{rl}{\dot{U}}_{R_a}& ={\dot{U}}_A-{\dot{U}}_{N^{\prime }}=(220-x\cos \theta )-j(x\sin \theta )\text{V}\\ \dot{I_{Ra}}& =\frac{\dot{U_{R_a}}}{R_a}=\frac{220-x\cos \theta -jx\sin \theta }{11}=\frac{440-2x\cos \theta }{22}+j\frac{-2x\sin \theta }{22}\text{A}\\ {\dot{U}}_{R_b}& ={\dot{U}}_B-{\dot{U}}_{N^{\prime }}=(-110-j110\sqrt{3})-(x\cos \theta +jx\sin \theta )\text{V}\\ \dot{I_{R_b}}& =\frac{\dot{U_{R_b}}}{R_b}=\frac{(-110-x\cos \theta )-j(x\sin \theta +110\sqrt{3})}{22}=\frac{-110-x\cos \theta }{22}+j\frac{-x\sin \theta -110\sqrt{3}}{22}\text{A}\\ {\dot{U}}_{R_c}& ={\dot{U}}_C-{\dot{U}}_{N^{\prime }}=(-110+j110\sqrt{3})-(x\cos \theta +jx\sin \theta )\text{V}\\ \dot{I_{R_c}}& =\frac{\dot{U_{R_c}}}{R_c}=\frac{(-110-x\cos \theta )+j(110\sqrt{3}-x\sin \theta )}{22}=\frac{-110-x\cos \theta }{22}+j\frac{-x\sin \theta +110\sqrt{3}}{22}\text{A}\end{array}$$

虚部相加等于零：
$-2x\sin \theta -x\sin \theta -110\sqrt{3}+110\sqrt{3}-x\sin \theta =0$
化简得 $x\sin \theta =0$
因 $x\ne 0$ V，故 $\theta =0^{\circ }$

实部相加等于零：
$440-2x\cos \theta -110-x\cos \theta -110-x\cos \theta =0$
化简得 $x\cos \theta =55$
因 $\theta =0^{\circ }$，故 $x=55$ V

$$\begin{array}{rl}{\dot{U}}_{R_a}& ={\dot{U}}_A-{\dot{U}}_{N^{\prime }}=165\angle 0^{\circ }\text{V}\\ {\dot{U}}_{R_b}& ={\dot{U}}_B-{\dot{U}}_{N^{\prime }}=-165-j110\sqrt{3}=252\angle {-130.9}^{\circ }\text{V}\\ {\dot{U}}_{R_c}& ={\dot{U}}_C-{\dot{U}}_{N^{\prime }}=-165+j110\sqrt{3}=252\angle {130.9}^{\circ }\text{V}\end{array}$$

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（3）无中性线且 A 相短路时

$$\begin{array}{rl}{\dot{U}}_{R_b}& ={\dot{U}}_B-{\dot{U}}_A=220\angle (-120^{\circ })-220\angle 0^{\circ }=-330-j110\sqrt{3}=380\angle (-150^{\circ })\text{V}\\ {\dot{U}}_{R_c}& ={\dot{U}}_C-{\dot{U}}_A=380\angle 150^{\circ }\text{V}\\ \dot{I_{R_b}}& =\frac{\dot{U_{R_b}}}{R_b}=\frac{380\angle (-150^{\circ })}{22}=17.3\angle (-150^{\circ })\text{A}\\ \dot{I_{R_c}}& =\frac{\dot{U_{R_c}}}{R_c}=\frac{380\angle 150^{\circ }}{22}=17.3\angle 150^{\circ }\text{A}\\ {\dot{I}}_{R_a}& =-({\dot{I}}_{R_b}+{\dot{I}}_{R_c})=30\angle 0^{\circ }\text{A}\end{array}$$

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（4）无中性线且 C 相断路时

$$\begin{array}{rl}\dot{I_{BA}}& =\frac{\dot{U_B}-\dot{U_A}}{R_A+R_B}=\frac{380\angle (-150^{\circ })}{11+22}=11.5\angle (-150^{\circ })\text{A}\\ \dot{U_{R_a}}& =\dot{I_{BA}}\cdot R_a=11.5\angle (-150^{\circ })\cdot 11=127\angle (-150^{\circ })\text{V}\\ \dot{U_{R_b}}& =\dot{I_{BA}}\cdot R_b=11.5\angle (-150^{\circ })\cdot 22=253\angle (-150^{\circ })\text{V}\end{array}$$

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（5）若有中性线时

• 情况（3）：A 相短路导致火线与中线短路（故障状态）
• 情况（4）：A、B 相电压电流保持正常（不受影响）