answersLogoWhite

0

If by 'loss current' (?) you mean the 'copper losses', then there is no practical way of doing so without reducing the load current. Perhaps you should explain what you understand 'loss current', in the context of a step-down transformer, to mean?

User Avatar

Wiki User

13y ago

What else can I help you with?

Related Questions

A single-phase transformer when supplied from 220v50hz has eddy current loss of 50wif the transformer is connected to a voltage of 330v50hzthe eddy current loss will be?

112.5


Is a device that changes the voltage of an alternating current with little loss of energy?

The device you are referring to is called a transformer. Not only can a transformer increase voltage, it can decrease voltage as well. These are known as a step up transformer for increasing voltages and step down for decreasing voltages.


Function of no-load in transformer in a open-circuit?

To calculate the no load current from transformer & core loss is also calculated.


What is exciting current in transformer?

Excitation current is the current necessary to "turn on" the transformer so it can be used. It's energy that is lost in the use of the transformer. Most of this loss I believe is associated with the hysterisis loop, although some will be lost as eddy currents.


No load current of transformer is non sinusoidal even though sinusoidal voltage is provided in oc sc test?

eddy current loss in the transformer core is reduced by


Why rated current should flow in the secondary for short circuit?

The short-circuit test runs the rated current through the windings of the transformer to measure the copper-loss, or power lost in the winding resistances. Because the transformer is run at low voltage for this test, there is very little additional power loss in the iron core of the transformer.


How does eddy current loss effect a transformer?

Eddy currents act to increase the temperature of a transformer's core above ambient temperature, resulting in a loss of energy through heat transfer -thus reducing its efficiency.


Why the efficiency of a power transformer increases with a decrease in Load at constant PF?

Maximum efficiency of a power transformer occurs when copper loss equals to iron losses. Decrease in current does not result in increase in efficiency unless the copper loss was more than iron loss and the decreased current made the copper loss is reduced and became equal to iron loss at some point.


How much to run a 75 KVA transformer with no load?

75 kV.A is the rated apparent power of the transformer which it can supply, continuously, to a load without overheating. When the transformer is not supplying a load, the primary current is (a) very small and, (b) lagging the supply voltage by practically 90 electrical degrees. Bear in mind that energy losses only occur for the component of current that is actually in phase with the supply voltage. So the energy consumed to the transformer is very small and is due to the resistance of the primary winding (copper loss) and a relatively small loss in the core (iron loss). Just how much energy this accounts for and, therefore, how much it costs to run the off-load transformer, is not possible to tell without knowing the full specification of the transformer.


How eddy current losses andhysterious losses can be reduced?

eddy current can be reduced by using laminated cores. and also be reducing the thickness of the stampings. transformer iron loss is the combination of eddy current loss and hysterisis loss. both the losses depend on core of the transformer and iron loss is a constant loss.


What is percentage of transformer loss at no load condition?

major component of power loss in a transformer is secondary resistance.when transformer is operated under no load,no current flows through the secondary.so under no load conditions transformer has just very small megnetic losses.


What does a step up transformer increase?

Power is normally transmitted at high voltage through step up/step down transformers to minimize the power losses in the transmission lines (this is one reason anyway). Since power loss is equivalent to the resistance of the conductor times the current squared, stepping up the voltage by a factor of two cuts the transmission losses by a factor of (2^2 = ) 4.