当故障正确发生时，会出现坏的情况
断路器自身的下游。从故障点看，电路为
由带变压器的网络系列表示。短路
电流与CB1相同。
电机的额定电流等于385 A；要选择的断路器
是Tmax T5H 400。
CB3的选择
对于CB3，当故障发生在正下游时，坏的情况也会出现
断路器本身。
从故障点看，电路由两个并联支路表示：
电机和系列的网络和变压器。根据
在前面的规则中，短路功率是通过使用以下公式确定的
公式：
电机//（网络+变压器）
负载L的额定电流等于1443 A；要选择的断路器
是Tmax T7S1600或Emax X1B1600。
示例2
图中所示的电路由电源、两个变压器组成
并联和三个负载。
上游网络：Ur1=20000 V
Sknet=500 MVA
变压器1和2:Sr
=1600 kVA
英国
% = 6%
U1r/U2r=20000/400
负载L1:Sr
=1500 kVA；cosν=0.9；
负载L2:Sr
=1000 kVA；cosν=0.9；
负载L3:Sr
=50 kVA；cosν=0.9。

the worst condition arises when the fault occurs right downstream of the circuit-breaker itself. The circuit, seen from the fault point, is represented by the series of the network with the transformer. The short-circuit current is the same used for CB1. The rated current of the motor is equal to 385 A; the circuit-breaker to select is a Tmax T5H 400. Selection of CB3 For CB3 too, the worst condition arises when the fault occurs right downstream of the circuit-breaker itself. The circuit, seen from the fault point, is represented by two branches in parallel: the motor and the series of the network and transformer. According to the previous rules, the short-circuit power is determined by using the following formula: Motor // (Network + Transformer) The rated current of the load L is equal to 1443 A; the circuit-breaker to select is a Tmax T7S1600 or an Emax X1B1600. Example 2 The circuit shown in the diagram is constituted by the supply, two transformers in parallel and three loads. Upstream network: Ur1=20000 V Sknet = 500 MVA Transformers 1 and 2: Sr = 1600 kVA uk % = 6% U1r /U2r =20000/400 Load L1: Sr = 1500 kVA; cosϕ = 0.9; Load L2: Sr = 1000 kVA; cosϕ = 0.9; Load L3: Sr = 50 kVA; cosϕ = 0.9.