Class X CTs are special CTs used mainly in balanced protection systems
(including restricted earth fault) where the system is sensitively dependent
on CT accuracy. Further to the general CT specifications, the manufacturer
needs to know:
• Vkp - Voltage knee point
• Io - Maximum magnetising current at Vkp
• Rs - Maximum resistance of the secondary winding
RCT is Secondary resistance of Current transformer it depends on Design.
don't worry
Tis question is incompleet.1000 va =1 kva.This is the power capacity of transformer. A transformer having 2 currents Primary current and secondary current . for that we required both voltage. Simply we can calculate by a formula Voltage x Current x 0.8(power factor)=1000.
Leakage reactance is useful for limiting the short circuit current in transformer and generators. therefore normally the reactance of transformer varies for 4.5% for distribution transformer to 12.5% for 400KV class.
Step 1: calculating knee point voltage Vkp Vkp = {2 x Ift (Rct+Rw)}/ k Vkp = required CT knee point voltage Ift = max transformer through fault in ampere Rct = CT secondary winding resistance in ohms Rw = loop impedance of pilot wire between CT and the K = CT transformation ratio Step 2: calculate Transformer through fault Ift Ift = (KVA x 1000)/(1.732 x V x Impedance) KVA = transformer rating in kVA V = transformer secondary voltage Impedance = transformer impedance Step 3: How to obtain Rct To measure when CT is produce Step 4: How to obtain Rw This is the resistance of the pilot wire used to connect the 5th class X CT at the transformer star point to the relay in the LV switchboard. Please obtain this data from the Electrical contractor or consultant. We provide a table to serve as a general guide on cable resistance. Example: Transformer Capacity : 2500kVA Transformer impedance : 6% Voltage system : 22kV / 415V 3phase 4 wire Current transformer ratio : 4000/5A Current transformer type : Class X PR10 Current transformer Vkp : 185V Current transformer Rct : 1.02½ (measured) Pilot wire resistance Rw : 25 meters using 6.0mm sq cable = 2 x 25 x 0.0032 = 0.16½ Ift = (kVA x 1000) / (1.732 x V x impedance) = (2500 x 1000) / (1.732 x 415 x 0.06) = 57,968 round up 58,000A Vkp = {2 x Ift (Rct+Rw) } / k = {2 x 58000 (1.02+0.16) } / 800 = 171.1½
These are the Class codes for CTs used for protection purposes.
CL of a CT is its accuracy class.. it is an approximate measure of the CT's accuracy. e.g. The ratio (primary to secondary current) error of a Class 1 (CL:1.0) CT is 1% at rated current
CL of a CT is its accuracy class.. it is an approximate measure of the CT's accuracy. e.g. The ratio (primary to secondary current) error of a Class 1 (CL:1.0) CT is 1% at rated current
In an ideal transformer, if the voltage is stepped up by a factor of x, then the current is stepped down by a factor of x. The end result is that the power, P=VI, is not changed. Again, this is in the ideal case.
in welding machine, the transformer used is basically a current transforrmer. so by changing the tap of secondary side of transformer, we can increase or decrease the current in the transformer.
why does have to short-circuit secondary wire of current transformer ?
Current = voltage x resistance. Therefore the current will be the same, assuming both frequencies are the same input voltage.