96%
That is the maximum efficiency occurs when the copper losses are equal to the core losses of the transformer.
torque in * input rpm/output rpm = torque out
Try at a electrical wholesale outlet. Don't know why you would want a auto transformer. A standard 240 to 120 transformer usually does the job and it can be run backwards or forwards. 240 primary 120 output on secondary or 120 primary 240 output on secondary. Either way you will have to know the amperage of the load to get the correct size of transformer.
Basically, you need to calculate the amount of energy contained in the fuel you use to produce a given amount of electrical energy. To calculate the amount of energy in the fuel you need to lookup the number of thermal energy units per unit volume then you can convert those thermal energy units into Joules (1 gallon [U.S.] of diesel oil = 146 520 000 joules - www.onlineconversion.com) The on the electrical side, just multiply the voltage by the current by the number of seconds of run time (unvarying load) to get the electrical power output in Joules. Then the efficiency is just the the output power divided by the input power (x100 for %).
Is 120 V the primary or secondary voltage? If the primary (input) voltage is 120, then at full load the transformer will draw about 0.42 A from the line, and the current delivered to the load depends on the secondary (output) voltage. If the secondary (output) voltage is 120, then at full load the transformer will deliver about 0.42 A to the load, and the current drawn fom the line depends on the primary (input) voltage. amps = watts / volts So, for instance, if your transformer has a 120 V primary, and a 24 V secondary, as you might find in your AC unit or furnace: Primary current (at full load) - A = W / V A = 50 / 120 A = 0.4166 Secondary current - A = 50 / 24 A = 2.08
A: a transformer will follow the rule of input output ratio with no load. As soon as a load is applied there will be changes in the ratio
It depends on the required output current, load rejection factor, and ripple. Also, efficiency enters into the picture.
the efficiency is maximum in a transformer when no load loss is equal to load loss.
For an 'ideal' transformer operating at full load, the answer is yes. But, 'real' transformers are a little less than 100% efficient so, in practice, the input power will slightly exceed the output power. In most circumstances, for the purpose of calculating primary and secondary currents, we can assume 100% efficiency.
The load side of a transformer feeds the device, such as a light or motor. It is the output of the transformer. The input, or line side, provides the voltage that is to be transformed, either up or down, to supply the load side.AnswerA transformer's primary winding is connected to the supply voltage, and the secondary winding is connected to the load.
A: Assuming 100% efficiency 320 ma
If the pulley is frictionless the angle does not matter, the tension in the rope will be the same on both sides of the pulley. If there is friction in the pulley however then you want to reduce the friction as much as possible. You do that by creating as large an angle between the two ropes as possible.
It is defined as dc power delivered to the load to the ac input power from secondary transformer
The change in output voltage from no load to full load defines the voltage regulation of that transformer.
The input of a load cell is the force or weight applied to it. The output of a load cell is an electrical signal, typically in the form of voltage or current, that is proportional to the applied force or weight.
For a motor's output power to equal its input power, the motor's efficiency must be 100%. As no machine, particularly a rotating machine, can possibly achieve 100% efficiency, there is no condition under which its output power can ever match its input power.
That is the maximum efficiency occurs when the copper losses are equal to the core losses of the transformer.