15ohm is transfered to the primary by the square of the windings ratio.

 

Therefore, on the primary it is now 15/10^2=0.15 which is very small against the 5ohn in series resulting in not much power transferred.

 

@uwed OK, but that 5 ohm resistance on the input side is wrong. I think that the resistor R1 takes ALL the voltage, since the coil has no given resistance - it's 0 ohm, isn't it? That makes it useless. But what if there is no resistor, but rather the coil has the resistance R1 = 5 ohm? Is that what you did for your calculations?

 

I just solved the current equations of the circuit you have drawn. I did not think much about what you wanted to model.

If the coil has a primary side windings resistance of 5ohm and zero ohm secondary side windings resistance the transformer will not transfer much power. But realistically, the secondary side windings resistance will not be zero but much larger because there are 10 times more windings (in your example).

 

I managed to solve my example and it seems to be ok.

I just had to calculate voltage drop on a primary coil itself to know the voltage trnasformed to the secondary coil. Now it all seems to make sense.

 

The power grid is based on the Faraday Transformer laws, the higher the voltage for transmission, the less voltage drop over a long distance, so transformers are use to step up the voltage at the power station, and then stepped down at the sub station to the customers,

whatever wattage the transformer is designed for the in the primary, the secondary is just less than this due to losses, so 10Kv @1amp= 100V@100amp,

 

@uwed OK, but what about that example? Now on the input side the coil has R1 = 5 ohm. I tried solving it, but something goes wrong:

 

(1) R1 is not shown as a discrete element in your circuit! Let's suppose it is like in your previous circuit. Then I1=10 (correct)
(2) Due to the voltage drop at R1 there would be just 100V on the secondary resulting in I2=5A.
(3) Input power is 1kW (correct). Output power is Pout=100V*5A=0.5kW. The missing 500W are lost at R1. Transformer efficiency is just 50% (compare to a MV-grid: 98.5%)

 

primary is 100V @ 20amp, so secondary is 200V@10amp.

 

... if R1=0 which is not clear from the circuit drawing.

 

Yeah, R1 would be 0. Now I see I did it wrong. I can't give a coil 5 ohms. I have to draw a resistor in series - it's a model.

If input and output both have 100 V, it isn't really needed to use a transformer. Or rather the transformer should be designed differently. What should I do to actually INCREASE the voltage on the output? Should the windings ratio be bigger? I think that adding resistance on the input side makes the whole problem even more problematic.

 

The resistances in your example are not realistic and that makes your calculations confusing. Realistically you design the transformer with windings cross sections so that the windings resistances at primary and secondary are very small as compared to the load resistance. Then reduction in transferred voltage will be very small and transformer efficiency will be close to 100%. Industrial distribution transformers in the MW-range have efficiencies around 98.5 - 99.0%.

 

So could you please give me an example of a well-designed step-up transformer with the proper resistances? I would do some calculations to see if it works.

 

how many times must you be told the same thing to understand? the only thing transformer resistances effect is how much current the transformer windings will handle.

 

Just make R1 << Rload/ratio^2

 

e.g. Rload=15ohm --> R1=0.05*15/10^2=7.5m Ohm

 

The primary and secondary windings are reading zero ohms at DC. Transformers don't work with DC, they work with AC. When you apply AC, you get an impedance (AC resistance) which is dependent on the number of windings and source/load resistance.

 

Getting back to you original question. If the current did not behave as it does then you would have a method of creating more power from less power and that violates several well known laws of Physics. Whenever you have two things multiplied together, like voltage and current, equaling a constant like power. You have this behavior. If you have a method of increasing one of the variables the other MUST decrease, and it must decrease in a particular way.

Other examples:
c = f⋅λ ⇒ speed of light = frequency time wavelength
P⋅V = n⋅R⋅T ⇒ Pressure times Volume = #of moles⋅Rydberg Constant⋅Absolute Temperature in °K
d = r⋅t ⇒ distance = rate times time

The graph of each of these relations is called a hyperbola.

My favorite curiosity is the horsepower.
One horsepower is 550 ft-lbf/sec
That means if I have a typical horse and he can lift 550 lbf., one ft. in one second he has exerted one horsepower. Now however if I have a magical horse who can lift 1 lbf. 550 ft. in one second he has also exerted one horsepower.

 

Let's look at it another way.

The grid is designed to deliver a certain amount of power, lets say 1MW.

Now, you can get 1MW by supplying 100V at 10000A.

Or, you can get 1MW by supplying 10000V at 100A.

Supplying 10000A will require 100 times as much copper as supplying 100A.

Bob

 

Guys, none of this is the problem that the TS was/is having. He was getting tripped up by the common misconception that the step/up and step/down relations apply to changes in the turns ratio. Namely, if a transformer with a particular turns ratio is providing a certain voltage at a certain current to a circuit, that if you double the turns ratio that the voltage will double and the current will be cut in half RELATIVE TO the voltage and current delivered prior to the change. The root cause of this misconception is not recognizing that the current drawn by the primary of the transformer will not remain the same when the turns ratio is changed.