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| Consider
an ideal transformer. In an ideal transformer due to no power loss, power input is equal to power output i.e. |
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|
Power
input = Power out put
Pp = Ps [power = VI] Vp Ip = Vs Is Vp/Vs = Is/Ip |
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| This expression indicates that current and voltage have an inverse relation with each other in transformer. | ||||
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EFFICIENCY
OF TRANSFORMER
|
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| Efficiency
of a device is equal to the ratio of output to input. Since, Output
= Ps Input
= Pp Efficiency
= output / input In actual
practice output is not equal to input therefore actual transformers are
not 100% efficient. However commercial transformers have very high efficiency
in the range of 95% to 99%. |
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POWER
LOSS IN TRANSFORMER
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| There are various factors that reduces the efficiency of a transformer such as: | ||||
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EDDY
CURRENTS
|
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| Due to variation in magnetic flux eddy currents are induced on the surface of iron core which in turn produce heating and therefore reduce the amount of power to the secondary coil. | ||||
| In order to avoid eddy currents , the core is laminated, made of thin sheets of soft iron. Each sheet is separated from the next by a layer of insulating varnish. | ||||
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HYSTERESIS
LOSS
|
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| Each time the direction of magnetization is reversed, some useful energy is wasted in overcoming internal friction. This is known as "hysteresis loss" and it also produces heating in the core. | ||||
| Hysteresis loss is minimized by using special alloys known as "perm alloy" for core material. | ||||
|
HEATING
|
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| Some energy is dissipated as heat in the coil. | ||||
| This is reduced by using suitablythick wire. | ||||
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FLUX
TRANSPORT FAILURE
|
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| Some loss of useful energy occurs because a small amount of the flux associated with the primary coil fails to pass through the secondary. | ||||
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