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There are basically 4 major differences :- 1. The windings (both primary and secondary) of an ideal transformer are considered to have zero resistance, hence the transformer is lossless. 2. There is no leakage flux in an ideal transformer. 3. The permiability of the core material in ideal transformer is considered to be tending to infinity and hence the current needed to set up the flux in the transformer is negligible. 4. There is zero hysterisis and eddy current losses in an ideal transformer.
Ideal transformer is useful in understanding the practical transformer..i does't have losses...
No. Available step current is inversely proportional to available step voltage. For example, if you have a turns ratio of 10:1 for a typical step-down transformer running off of 120 VAC, producing 12 VAC; if the input current were 1 ampere, the output current would be 10 amperes. Similarly, for a step up transformer, available voltage goes up while available current goes down, all within the turns ratio. Nope. The current will be equal if the turns ratio is 1:1 in an ideal transformer. But, t/f s are not designed that way. Further, Current ratio is equal to the inverse of turns ratio.
A: Transformer by itself goes not alter the phases the output can be in phase or out of phase depending how you look at it.
In an ideal transformer the power in equals the power out there is no gain. In an ideal amplifier the power out equals the Gain *Power In. An ideal transformer transforms energy at a ratio of its windings. For example an ideal 1:10 ratio transformer (step up) would convert a 10Volt input at 10amps to a 100Volt output at 1Amp. Or conversely an ideal 10:1 ratio transformer (step down) would convert a 10Volt 10Amp input into a 1Volt 100Amp output. Since Power = Voltage * Current we can see the power in equals the power out in an ideal transformer. In an ideal amplifier the power out is greater than the power in. This is defined as the gain of the amplifier. An ideal amplifier with a Voltage gain of 10 would take a 1Volt 1amp signal and amplify it into a 10Volt 1 amp signal. An ideal amplifier with a Current gain of 10 would take a 1Volt 1Amp signal and amplify it into a 1Volt 10 amp signal. Since Power = Voltage * current we can see the power in is less than the power out showing a gain in power.
There are basically 4 major differences :- 1. The windings (both primary and secondary) of an ideal transformer are considered to have zero resistance, hence the transformer is lossless. 2. There is no leakage flux in an ideal transformer. 3. The permiability of the core material in ideal transformer is considered to be tending to infinity and hence the current needed to set up the flux in the transformer is negligible. 4. There is zero hysterisis and eddy current losses in an ideal transformer.
In low voltage and electronics Leakage Current is any current that flows when the ideal current
In an ideal transformer, if the voltage is stepped up by a factor of x, then the current is stepped down by a factor of x. The end result is that the power, P=VI, is not changed. Again, this is in the ideal case.
There are basically 4 major differences :- 1. The windings (both primary and secondary) of an ideal transformer are considered to have zero resistance, hence the transformer is lossless. 2. There is no leakage flux in an ideal transformer. 3. The permiability of the core material in ideal transformer is considered to be tending to infinity and hence the current needed to set up the flux in the transformer is negligible. 4. There is zero hysterisis and eddy current losses in an ideal transformer.
Ideal transformer is useful in understanding the practical transformer..i does't have losses...
ideal transformer is that which has no power losses.if any transformer transfer power to secondary without power loss then that call a ideal transformer
No. Available step current is inversely proportional to available step voltage. For example, if you have a turns ratio of 10:1 for a typical step-down transformer running off of 120 VAC, producing 12 VAC; if the input current were 1 ampere, the output current would be 10 amperes. Similarly, for a step up transformer, available voltage goes up while available current goes down, all within the turns ratio. Nope. The current will be equal if the turns ratio is 1:1 in an ideal transformer. But, t/f s are not designed that way. Further, Current ratio is equal to the inverse of turns ratio.
In an ideal xmfr, the power in equals power out and only the voltage and current changes.
A: Transformer by itself goes not alter the phases the output can be in phase or out of phase depending how you look at it.
Ideal gas law. At a fixed temperature, the pressure and volume are inversely related. PV=mRT
In an ideal transformer the power in equals the power out there is no gain. In an ideal amplifier the power out equals the Gain *Power In. An ideal transformer transforms energy at a ratio of its windings. For example an ideal 1:10 ratio transformer (step up) would convert a 10Volt input at 10amps to a 100Volt output at 1Amp. Or conversely an ideal 10:1 ratio transformer (step down) would convert a 10Volt 10Amp input into a 1Volt 100Amp output. Since Power = Voltage * Current we can see the power in equals the power out in an ideal transformer. In an ideal amplifier the power out is greater than the power in. This is defined as the gain of the amplifier. An ideal amplifier with a Voltage gain of 10 would take a 1Volt 1amp signal and amplify it into a 10Volt 1 amp signal. An ideal amplifier with a Current gain of 10 would take a 1Volt 1Amp signal and amplify it into a 1Volt 10 amp signal. Since Power = Voltage * current we can see the power in is less than the power out showing a gain in power.
with an ideal electrical transformer with an input current of 2 amps and an output current of 1 amp what is the turns ratio of the secondary and primary coils