We all know Ohm's law and power conservation, but in a basic transformer, they seem to be contradictory. For example, say we have a 50v source running through a 10Ω wire, then the amperage to the transformer should be 5 A and would come out to 250 watts. Then say the transformer steps the voltage down to 5v, then according to power conservation the amperage should be 50 A to come out to 250 watts. Here's where my problem comes in -- Ohm's law tells us I=V/R, therefore if we assume the wire the output of the transformer has also has a resistance of 10Ω then the Amperage would be .5 Amps. This doesn't really make any sense to me, any help is appreciated.
The resistance of the second circuit would affect how much current flows in the first one. Also, you are neglecting losses. P_{in} - Losses = P_{out} This is where the .5 Amps comes in. That is current that is lost to heat in the second wire from the resistance. One more thing, this is an AC circuit, so you should be talking about impedance, not just resistance. edit: I should note, that had I done the exact math, the .5 amps wouldn't be correct, but I am lazy and I am not going to do it. But IF voltage was 5v, and IF 50A was being forced through the line, then that 10 ohm wire would be losing .5 amps to heat and not limiting the current to .5 amps.
Ohms Law only applies to DC, and only applies to "Ohmic conductors" - i.e. those which obey Ohm's Law. I know that's circular but you can look it up. It's a bit like how Hook's Law only applies below the "elastic limit"... It does not apply to AC (see "reactance" and "impedence"). It does not apply to transformers (see "mutual inductance").
When you talk about a transformer dissipating 250 watts in a 10 ohm primary. that is 250 watts of lost power (dissipated as heat, that is). For example, say it was 50V DC. The current is limited by the resistance. An ideal transformer has no resistance in the primary or secondary. The primary impedance is a reflection of the secondary load. There is no loss in the transformer. If you applied 50V DC you would get infinite current (ignoring transient response). Say you put a 1 ohm load on the secondary of a 10:1 transformer and applied 50V AC to the primary. On the secondary,5 volts, 1 ohm gives 25 watts in the resistor. This load would appear as 100 ohms on the primary so 50V into 100 ohms give 25 watts input into the transformer, but not dissipated in the transformer. All the power is dissipated in the secondary load. Read the wikipedia transformer page.
That's debatable. It has the form of ohm's law, but it is not ohm's law. From wikipedia, for example "When reactive elements such as capacitors, inductors, or transmission lines are involved in a circuit to which AC or time-varying voltage or current is applied, the relationship between voltage and current becomes the solution to a differential equation, so Ohm's law (as defined above) does not directly apply since that form contains only resistances having value R, not complex impedances which may contain capacitance ("C") or inductance ("L"). Equations for time-invariant AC circuits take the same form as Ohm's law, however, the variables are generalized to complex numbers and the current and voltage waveforms are complex exponentials." also: "Materials and components that obey Ohm's law are described as "ohmic" [27] which means they produce the same value for resistance (R = V/I) regardless of the value of V or I which is applied and whether the applied voltage or current is DC (direct current) of either positive or negative polarity or AC (alternating current)." But saying that ohms law didn't apply in the OP's problem was of no value and had nothing to do with what he was actually confused about. None of his problem was related to a misapplication of ohm's law to a non-ohmic device.
first of all that's how it is made. besides voltage and current, there are other factors working here. If the primary side of the transformer is connected to such a source that can push 5A on 50V terminal that means u can construct transformer that can provide upto 250W on secondary(neglecting all losses). how much current and voltage u will get on ur secondary depends on the transformer construction. u can make a 5A 1A = 5W transformer or u can make a 5V 10A = 50W Transformer. (although as it is AC voltage we are talking about ,transformers are rated at VA not W) so the thing is transformer will only take up such amount of power that its supposed to take. there is available power at the primary side doesn't mean it will take up all that power. Losses: 1. copper loss: transformers are made of copper wire so when ever electric current flow through them they emit energy in the form of heat. thus energy is wasted here. 2. core loss: it supplies current for hysteresis and eddy loss. 3. Magnetizing loss: supplies for magnetically linking up primary and secondary windings.
There is no contradiction if you apply the definition of Resistance in the right place. The load resistance determines the current that will flow due to the secondary volts. The transformer transforms the current ratio between I2 ands I1 so that the generator 'sees' a transformed version of the load with a different value of primary current. This means that I1V1 =I2V2 (ignoring any efficiency factor). The idea of a transformer transforming a resistance takes a bit of effort to accept at first but it's the best way of looking at what happens. (We've all heard about 'Matching transformers', and they are used for just that purpose.) V1/V2 = N1/N2 (Relationship between volts and number of turns) I1/I2 = N2/N1 (Relationship between current and number of turns) and R_{in}/R_{load} =(N2/N1)^{2} (R_{in)} is what the generator sees)
I've also heard the one that "a motor and a transformer are virtually the same". Just the "secondary" of the motor is free to spin.