Why will current decrease in a step-down transformer ?

[Deathnote777]

Why will current decrease when voltage increase in a transformer ?

I 've just learned it and I get confused. Assume there are 2 solenoids, the no. of turns of coils in the left sides one(with AC current) is less than that in right sides.By conservation of energy, V(left) x I(left) = V(right) x I(right). It is seemingly correct. Now, we consider only the right side one, assume V(left) x I(left) is constant. If the no. of coils of the right-side solenoid is doubled, the emf(voltage) induced will increase, as well as the current (I). However, the current will drop by conservation of energy. How can you explain why transformer current is inversely proportional to voltage ?
Would it have any difference if I replace the left-side solenoid with a rotating magnet? Thx ~

http://imageshack.us/photo/my-images/217/img20120725153619.jpg/

http://imageshack.us/photo/my-images/402/img20120725153931.jpg/ <-- will the current increased when i double the no. of turn of coils ?

 

[willem2]

I 've just learned it and I get confused. Assume there are 2 solenoids, the no. of turns of coils in the left sides one(with AC current) is less than that in right sides.By conservation of energy, V(left) x I(left) = V(right) x I(right). It is seemingly correct. Now, we consider only the right side one, assume V(left) x I(left) is constant.

And what will you do to make V(left) x I(left) constant?

If you connect the primary to a good voltage source, and the secondary to a resistance R, if you double the coils on the secondary, you'll get twice voltage across R, so twice the current through R, but the current through the primary will also increase.

 

[papernuke]

If the no. of coils of the right-side solenoid is doubled, the emf(voltage) induced will increase, as well as the current (I). The induced current should be less than in the primary because it's a step-up transformer. There's more coils (and a higher voltage) in the right side, so the right side must have a lower current to conserve energy.

Also, what do you mean by a rotating magnet?

 

[Deathnote777]

And what will you do to make V(left) x I(left) constant?

If you connect the primary to a good voltage source, and the secondary to a resistance R, if you double the coils on the secondary, you'll get twice voltage across R, so twice the current through R, but the current through the primary will also increase.

I mean I won't change any factors in the left-side solenoid, but increase the no. of turns of coils in the right.

 

[Deathnote777]

Also, what do you mean by a rotating magnet?

I know the current will decrease to conserve energy, but if I consider only the right-side solenoid, it will be strange that when V rises, I drops. It contradict V = IR. For the rotating magnet, I want to assume there is a changing B-field caused by magnet, just like the field created by the solenoid. I added pic

 

[willem2]

I mean I won't change any factors in the left-side solenoid, but increase the no. of turns of coils in the right.

But the number of coils to the right does influence the current in the primary coil. A transformer works both ways. With twice as many turns, you get twice the voltage across the load of the secondary, and this will result in 4 times the flux change per unit time in the transformer from the secondary current, because the secondary has now twice the current AND twice as many turns, and to cancel that you need 4 times the current in the primary.

 

[Deathnote777]

But the number of coils to the right does influence the current in the primary coil. A transformer works both ways. With twice as many turns, you get twice the voltage across the load of the secondary, and this will result in 4 times the flux change per unit time in the transformer from the secondary current, because the secondary has now twice the current AND twice as many turns, and to cancel that you need 4 times the current in the primary.

Huh? The fact is, when the turns is doubled, V is doubled and the current in the secondary will be halved. What i get confused is why the current is halved when voltage is doubled

 

[willem2]

Huh? The fact is, when the turns is doubled, V is doubled and the current in the secondary will be halved. What i get confused is why the current is halved when voltage is doubled

But the current won''t be halved. The current will in the secondary will double and the current in the primary will go up by a factor of 4. You can't fix both v(primary) and I(primary). If v(primary) is fixed because you hook up the primary to a voltage source (like the mains), then I(primary) is determined by the load resistance and the turns ratio.

 

[Dale]

I know the current will decrease to conserve energy, but if I consider only the right-side solenoid, it will be strange that when V rises, I drops. It contradict V = IR. For the rotating magnet, I want to assume there is a changing B-field caused by magnet, just like the field created by the solenoid. I added pic

V=IR has very little to do with how a transformer functions, which is primarily via induction. You need to use the inductance formula: $v=L\frac{di}{dt}$ and the mutual inductance formula: $v_1=M\frac{di_2}{dt}$

 

[Deathnote777]

V=IR has very little to do with how a transformer functions, which is primarily via induction. You need to use the inductance formula: $v=L\frac{di}{dt}$ and the mutual inductance formula: $v_1=M\frac{di_2}{dt}$

Oh~I haven't learn that...Do you mean ohm's law is not working in transformer? And can you simply explain what is the formula about ? e.g. the R increase with voltage

 

[sophiecentaur]

I think it's important to get your 'cause and effect' s in the right order. It's a bit of a 'pulling up with your own bootstraps' argument. First, assume that the self inductance of the primary is high enough to make the off-load current negligible.
Every turn of the primary is producing magnetic flux and that flux is due to the current through it. The primary current is governed by the difference between the supply volts and the 'total / effective back emf'. The volts induced in the secondary are governed by the flux and the number of secondary turns. If the load takes a given current, then that current governs the amount of effective back emf (reducing it from its original value which is equal to the supply volts) that turns up in the primary and, hence, the primary current. The induced voltage in the secondary is the turns ratio times the input volts. The induced volts in the primary due to the secondary current is in phase with the supply volts and is secondary current divided by the ratio (reducing the back emf that the source will 'see'). This means that VI in each winding is the same.

There is a good mechanical analogy in a (mass-less) lever with a load on one end. The force applied to the input, produces a (scaled) movement on the output and the acceleration of the load produces a reaction against the output , which (also scaled) resists the input force. Force times distance (work) is the same on each side.

 

[Dale]

Oh~I haven't learn that.

Hmm. I don't want to over-do an explanation. Can you give me some background? I.e. have you dealt with complex impedance, do you know Kirchoff's voltage and current law, have you seen loop current or node voltage analysis, etc.