推荐回答（5个）
2500KVA的变压器,如果不投电容器，可带负荷最大只有2500*0.8=2000KW。
这时的有功负荷是2000KW，功率因数0.8，无功负荷是1500KW。
如果电容器全部投入运行，则无功负荷降至1500-2*10*30=900KVAR。功率因数提高到了0.91，能带负荷最大可以提高到2500*0.91=2275KW。
考虑到变压器只有在90%负载率时效率最高，应将可用负荷降低约10%
简单算法补偿变压器容量的20%--30%，但是我一般补偿30%，大一点比较好。例如变压器500KVA，那么变压器的电容补偿为500*0.3=150Kvar，即补偿150Kvar即可。
2500/0.4/1.732=3608A 是最大电流
2500*0.85的功率因数=2125KW就是最大功率
最大功率P=根号3*U*I*功率因数=1.732*400*3608*0.85=2125KW
电容补偿计算
一可以通过以下公式计算所需补偿容量：
公式一：Qc=p（√12φ－cos1/√ cosφ- φ－cos12/cosφ）
=p(tgφ1－tgφ)(kvar)
公式二：Qc=P（√－1φcos/√112 －1 φcos/12）
=P（tgφ1-tgφ）（推荐用公式二，计算量小）
式中cosφ1: 补偿前的功率因数
cosφ：你想达到的功率因数
Qc: 所需补偿容量（kvar）
p: 线路总功率（KW）
补偿前的功率因数可以用功率因数表测量，或通过用电量进行计算： 功率因数＝有功电度／根号 有功电度平方＋无功电度平方
因为有时候，负荷的大小和功率因数确实不好准确计算，所以一般的设计部门就根据变压器的额定容量补30%--40%
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