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In the case of a transformer, why does increasing the voltage decrease the current 
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#1
Mar1112, 06:15 PM

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I mean,
If an ideal step up transformer increases the voltage from 10V to 20V say, why does the current halve? I know it happens and I know the equations for it, I know to say otherwise would contradict the law that energy cannot be created. But putting the equations aside, why exactly does it happen? EDIT: The part I don't understand is that when the potential difference increases, and the resistance (in the secondary) remains constant (I think?), why does current decrease? 


#2
Mar1112, 06:21 PM

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Because (minus losses) power is constant. Power = current * voltage.
Specifically 1 watt = 1 ampere * 1 volt. In more detail: For AC circuits the mean power is the average over a cycle of the voltage and current at a given time. This last detail is sometimes ignored in people inventing overunity devices with magnets and transformers and such. They miscalculate output power by multiplying peak voltage (occurring at one time) with peak current (occurring at a different time). 


#3
Mar1112, 06:42 PM

P: 18

I have been thinking more and I think (correct me if I'm wrong, please) that as there are more turns on the secondary, there is a greater load on the primary which means that the current through the primary would be reduced since the electrons travel slower. Since the magnetic field created is proportional to the current in the first place, the magnetic field is... smaller/less powerful? But here is where my understanding breaks down, so the voltage created (per turn) is still the same because...? And current decreases because...the magnetic field is smaller? 


#4
Mar1112, 07:57 PM

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In the case of a transformer, why does increasing the voltage decrease the current
Dave 


#5
Mar1112, 08:07 PM

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Say you have a 1turn coil with current I and voltage V. Let's make the wire multistranded with N strands. The strands are connected in parallel, each strand carries current I/N. Now connect the strands in series instead of parallel, making it into Nturn coil. Current through each strand and voltage around each strand remains the same, but current through the coil is now I/N and voltage across the coil V*N.



#6
Mar1112, 09:51 PM

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Energy in is energy out. The cause of the dynamic effects of a transformer are intimately related to the transformation of the energy into a different form without changing its amount. To understand why quantity A changes in a certain way you are going to have to see what that quantity means and quantities mean something only in equations. 1+1=2 gives 2 its meaning. Voltage has fundamental meaning as an electromotive force which means the work done by moving an amount of charge. You cannot be said to understand anything about voltage until you understand the equation Work done on/by a charge is (=) voltage times the amount of charge. Rate of work = power = energy per time = voltage times rate of flow of charge: Power = Voltage times Current. Until and unless you understand these basics, I don't see any merit into going through Maxwell's equations and talking about time rate of change of magnetic flux through a coil of wire, mutual and self inductances, and so on. The equation I gave IS the why and how. If you want to know the why and how of the output voltage (or current) relative to the input voltage (or current) you're going to have to understand this point first. Understanding must build upon understanding. 


#7
Mar1112, 10:06 PM

P: 18

The problem I have is that although I have understood the equations, I don't really understand the theory behind them. Its a bit like I know 1+1=2 but I don't know why its so. So I take a different approach, break down what really happens until I understand it. 


#8
Mar1112, 11:01 PM

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Around any flowing current is generated a magnetic field. Magnetic fields have energy proportional to the square of the field strength added up over the volume. This means a few things. One it means magnetic fields tend to spread out in order to minimize energy. It also means you can get more energy per current by concentrating the B field. This occurs when we loop the wire in a coil. The relationship between the energy of the B field around a coil of wire and the current through the wire is its self inductance. It has an analogous relationship to mass for a moving object (current being moving charge). We see this in the analog equations: Kinetic energy is 1/2 mass times velocity squared, Magnetic energy is 1/2 inductance times current squared. So if you have a coil of wire and you at one instant start to apply a voltage the current will begin to grow. The voltage is pushing the current through the wire. But there is also that power = voltage times current. That energy must go somewhere. With a resistor it goes into heat but if there is no resistance then all that energy will go into an ever growing magnetic field. It is analogous to a top spinning faster and faster as a constant torque is applied. Now you are applying a voltage to the ends of the looped wire so the voltage must change from one value to the other as you measure it along the length of the wire. What happens is that the growing B field around the wire will cause a back reaction, an induced voltage in the wire opposing your applied voltage. In the mechanical analogue imagine pushing a train (on a frictionless rail system). The whole train will accelerate together but each car feels only one part of the whole force you apply. The inertia of each car "pushes back" so that your whole force is spread out along each car. Just so the growing B field pushes back on the current in the wire. Suppose you apply a voltage for a time and the B field grows and then you remove the voltage? We can't just disconnect the voltage source as that will stop the current too. The energy stored in the B field cannot just disappear. What happens in real situations is that your attempt to disconnect doesn't work instantly, instead the "momentum" of the current keeps the charge flowing and it arcs through the air (at high resistance) causing the energy to be dissipated as heat in the spark. To avoid sparking but to reduce the current and B field back down to zero you can apply a reversed voltage for a time. The fact that the current flows against the voltage means negative work is done on the coil and rather the coil does work on the voltage source. If that is a battery it acts to recharge it. Now let's see what happens if you put two coils together, let's say in a 1 to 2 ratio. The first coil creates a certain amount of B field for a given current. The 2nd coil will produce twice as much B field and twice as much energy for the same current. But if you put a current of 2I in the first coil and a reverse current of I in the 2nd coil you get a net zero B field. Now try this. If you build up a current of 2I in the first coil, leaving the 2nd one open so no current flows, then once the field is built up you connect the 2nd coil and disconnect the first here is what will happen. (Again assuming 0 resistance). Instead of a spark the B field is able to keep going by inducing a current of 1I in the 2nd coil. It only needs to be half a great because it has twice the windings. You can then drain off the energy in the B field using the 2nd coil and a reverse voltage. In this way the double coil system can transfer energy to and from the B field, using either winding. The voltages can be anything you want, higher voltages just speed up the process. Also the currents in both coils can be arbitrary. Their effects add or cancel in so far as the B field is concerned depending on their direction. But what must be true is that as power goes in the same power goes out. Also what must be true is that the changing B field must have twice the effect on the doubly wound coil in terms of reaction voltage since it is related to half the current and power in = power out. Putting this all together and doing a bit of calculus with equations for voltage and current gives you a voltage ratio equal to the winding ratio and current ratio reversed and power in = power out. A very final note is that the above assumes the wires are so very small that all the B field loops through all the turns of both wires with no leakage between windings. This is not the case and a true 2:1 transformer won't quite give you 2:1 voltage and 1:2 current. The magnetic core helps but there is a loss of ratio. In addition resistance causes loss of efficiency (energy goes into heat instead of into B field or out of the secondary). 


#9
Mar1112, 11:21 PM

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#10
Mar1212, 01:48 PM

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Perhaps a better way round to think about it is this. It's a backwards way round argument but valid. It applies to an 'ideal transformer, which is just Fine to start with.
Assume the transformer is working and connected to a load R. Apply the Secondary Voltage V_{s} to Load Resistance R. The current I_{s} will be V_{s}/R. The secondary current is defined by this secondary load: no load (R = ∞), no current because of the self inductance of the Primary  a back emf. You can't FORCE current into the load except as a result of the secondary volts. V_{s} is related to V_{p} by the transformer ratio equation (with which we are more happy?) The Secondary current I_{s} will induce a voltage in the Primary which limits the amount of current in the Primary. So how much current will be 'allowed' into the Primary Winding? In the case of a 2:1 ratio transformer (or any ratio), the same power will have to flow in as flows out (or it will burn up or freeze down). The current (I_{p}) flowing into the supply terminals, with V_{p} across them stabilise out at half the value of I_{2}. As I_{s} increases (lower R) more current can flow in the the Primary until the current ratio is still 1:2. 


#11
Mar1312, 09:40 AM

P: 409

An important point about an ideal transformer is that the reluctance of the core is zero, this means in the magnetic circutit the sum of the magnetomotive force ( ƩN*I) must be zero otherwise the flux would be infinite. This results in the relation
I2/I1=N1/N2 And the ration happens to be the inverse of the voltage ratio. 


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