KV, kilovolts, or kilojoules per coulomb is not the same thing as KVA, kilovoltamperes, or kilojoules per second, and no direct comparison exists. Please restate the question.
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1)may be voltmeter or clipon meter will be connected across the winding. 2)or in motor winding we must find amp current and aramture resistance nd then find the ratings 3)formula is rated kva of motor*1000/base kva for transmission line.
kvar can be calculated as follows the a product KVA andt the sine of the angle between the KVA and KW.
it is depends on the rating of distribution transformer,e.g if there is 100 kva, then it will take 5.25amp.
3 phase kVA = V*I*sqrt(3) Where voltage is line to line, and current is the actual RMS current flowing in the a wire. kW = V*I*sqrt(3)*Cos (phi), where phi is the angle between the voltage and current; Cos (phi) is also known as the power factor. kVA is the vector sum of kW (real power) and kVAR (reactive power). As the equations above suggest, you must know the voltage to correctly calculate the current.
There isn't enough information here. Available short circuit fault level can be given as a KVA value for different types of faults, but I assume the questioner is looking for a relationshiop between (transformer?) KVA and available short circuit current - If my assumption is correct, there is no direct correlation without knowing the transformer positive and zero sequence impedances. If these are known, you can assume the source impedance is infinite, and calculate the maximum short circuit current through the transformer as follows: lowside fault current for a 3 phase fault on the lowside of the transformer: lowside kV (line to line) / (1.732 x per unit positive sequence impedance x scalar to real impedance), where scalar to real impedance is equivalent to lowside kV (line to line) ^2 / base kVA. For a L-G fault, do the same with zero sequence impedance.