No Confusion of √3

While modeling the balanced three phase system it is sufficient to draw single line diagram considering only one phase (excluding neutral). Thus while calculating pu values of elements for such (1-Ph) representation of the power system a question may arise in mind that whether to use 1-Ph values for base kV. However as the pu impedance value of element is its impedance multiplied by ratio of base MVA to square of base kV. Thus for 1-Ph calculations when you use 1-Ph base kV its value will be  times less and when squared this value will be 3 times less. While calculating 1-Ph pu value we have to consider 1-Ph MVA which is also 3 times less. Now as numerator (base MVA) and denominator (base kV) both are 3 times less the ratio and pu value will be same. Thus it is usual to use 3-Ph MVA and Line-Line voltage while calculating pu impedance values of elements

Same thing holds well while calculating 1-Ph short circuit currents. Here question may arise in mind; whether to use same conversion factors of current to convert 1-ϕ short circuit MVA to 1-ϕ short circuit currents. Instead of describing the fact in words it is described by an example. As shown in fig- consider a 50km 132kV line connecting source Bus-A and Bus-B; source impedance considered as zero (in academic language Bus-A is infinite bus) and SLG fault considered at Bus-B.

As shown in example calculations (See Fig- Part-3) to convert short circuit MVA to corresponding current we have to multiply short circuit MVA by 4.375 and not by 4.375/√3. This indicates simplicity of representing fault levels in MVA