500Mw at 11kv works out to 45000 a
a bit big for any wire manufactured on this planet
suggest you go to a more practical voltage like 132,000V
AnswerSince we are inevitably talking about a three-phase transformer, the line current is determined from
IL= rated apparent power / (1.732 x UL) = (500 x 106) / 1.732 x 11 x 103)
...which works out at 26 244 A which, as the original answer suggests, is not a practical value for this combination.
The product of the secondary rated current and the secondary rated voltage will give you the rated V.A of the transformer.
A DC voltage added to one side of a transformer has no effect on the other side.
Transformer rating is based on the maximum temperature a transformer can run at. This temperature is dictated by the amount of current flowing through the transformer windings. This is why transformers are rated in KVA (voltage * current), not kW - it doesn't matter what the phase relationship is between voltage and current, just the magnitude of the current.
copper loss is directly propostional to I (AMPERE) and iron loss directly propostional to V (VOLTAGE) then total losses is equal to volt ampere hence the rating of transformer in KVA. SULTAN
calculate the total sum of lighting power and add 10% as safe operation of the transformer. then select a transformer for that rating. suppose your total lighting power is 300VA. add 10% of the total lighting power so the total power is 330VA. therefore your transformer size can be 330va or more. please follow IEE regulations.
The product of the secondary rated current and the secondary rated voltage will give you the rated V.A of the transformer.
A transformer can be used to change the voltage to an appliance. The voltage rating of the transformer should be right for the voltages used, and the current rating of the transformer should not be less than the current drawn by the equipment.
The amp rating for a 100VA transformer will vary depending on the actual voltage of the transformer. Transformers have both a primary and a secondary voltage.
The secondary winding's current rating is the rated apparent power of the transformer (expressed in volt amperes) divided by its voltage rating (expressed in volts). This applies to both step down, and step up, transformers.
A DC voltage added to one side of a transformer has no effect on the other side.
A megger would not be suitable for testing insulation resistance of a 13.2-kV transformer, as the transformer's voltage rating is significantly higher than the output voltage from a megger.
Transformer rating is based on the maximum temperature a transformer can run at. This temperature is dictated by the amount of current flowing through the transformer windings. This is why transformers are rated in KVA (voltage * current), not kW - it doesn't matter what the phase relationship is between voltage and current, just the magnitude of the current.
Assuming that the voltage rating of the lamp matches the rated secondary voltage of the transformer, the lamp will operate at its rated power.
copper loss is directly propostional to I (AMPERE) and iron loss directly propostional to V (VOLTAGE) then total losses is equal to volt ampere hence the rating of transformer in KVA. SULTAN
A transformer is a power source. It will provide voltage to a device. Find the voltage rating on the device, say 24V. 250/24 = ~10A.
calculate the total sum of lighting power and add 10% as safe operation of the transformer. then select a transformer for that rating. suppose your total lighting power is 300VA. add 10% of the total lighting power so the total power is 330VA. therefore your transformer size can be 330va or more. please follow IEE regulations.
A CVT (Constant Voltage Transformer) is always better than a Servo Stabilizer. But a CVT is costlier than a Servo Stabilizer of same rating.