When transformers decrease voltage, do they increase the current?

[partialfracti]

Before a few days ago I always thought that when transformers decrease the voltage, the transformers increase the current. Then on another thread at Physics Forums, people have told me that when transformers decrease voltage, they don't increase the current. The PFers told me that the transformers increase the current available.

I think I understand it now.

Here's my current theory: When transformers decrease the voltage, the transformers increase the current that is available if a circuit is formed. Before a circuit is formed, there is no current. Is my current theory correct?

 

[vk6kro]

Yes, you have it right.

If no power is drawn from the lower voltage, then the current is only what is wasted in the transformer.

IF YOU DRAW THE SAME POWER AT A LOWER VOLTAGE... then the current must be greater because power = Voltage times current.

 

[yungman]

Current capability and voltage ratio of a transformer is kind of independent. YOu can only look at the rating of the transformer to determine how much you can draw.

If you are talking about the current ratio, if you have a 10:1 step down transformer, if you draw 1A on the secondary, you draw only about 0.1A at the primary.

 

[Antiphon]

This is confusing and wrong.

Put a load on a transformer and energize the primary.

Vprimary*Iprimary = Vsecondary*Isecondary.

If the transformer reduces the voltage by half it will exactly double the current. This is the definition of the ideal transformer.

 

[yungman]

This is confusing and wrong.

Put a load on a transformer and energize the primary.

Vprimary*Iprimary = Vsecondary*Isecondary.
Ideal transformer has core that has zero reluctance and the wires has infinite conductance!
If the transformer reduces the voltage by half it will exactly double the current. This is the definition of the ideal transformer.

We are talking about real life transformers. The wires has to support the increase in current. Current capability is limited by the size of the wire and the size of the core. You push it, the reliability will suffer. Yes, if you only talk about ideal transformer, everything is possible.

Of cause, if you just doing an experiment only and you'll be there while you are runing it. You can push how ever which way, turn it off when you start to smell something. But if you are designing a product, this is a no no to play this kind of pushing the limit if it is for a product.

 

[Antiphon]

We are talking about real life transformers. The wires has to support the increase in current. Current capability is limited by the size of the wire and the size of the core. You push it, the reliability will suffer. Yes, if you only talk about ideal transformer, everything is possible.

Of cause, if you just doing an experiment only and you'll be there while you are runing it. You can push how ever which way, turn it off when you start to smell something. But if you are designing a product, this is a no no to play this kind of pushing the limit if it is for a product.

You may be talking about real life transformers. The original post is not.

A real transformer is a very good approximation of an ideal one when it's loaded.

I have one in my laboratory that's rated for 100va. Its a 2:1 step down made by a real-world company that has no students designing it's products.

When I connect a 50w, 60volt lamp to the secondary and connect the primary to 120V, the primary current is HALF the secondary current, and the primary voltage is TWICE the secondary voltage to within 2%. The OP is asking about the 98% that I'm talking about, not the 2% that you're talking about.

 

[yungman]

I think the original question is confusing. I interpreted he meant if he reduce the voltage than the rated input voltage, the output voltage decrease and he can get more current out of it because it lower the power. I think what he meant is actually simple step down transformer!

I modified my original post. If you meant when you draw 1A on the secondary of a 10:1 transformer, then you draw about 0.1A at the primary. If that is what he meant.

 

[partialfracti]

If you are talking about the current ratio, if you have a 10:1 step down transformer, if you draw 1A on the secondary, you draw only about 0.1A at the primary.

Then why do you say that the current capability and voltage ratio of a transformer is kind of independent?

 

[partialfracti]

If the transformer reduces the voltage by half it will exactly double the current.

If the transformer reduces the voltage by half it will exactly the double the current when there is a load on the transformer. Is this correct?

 

[Antiphon]

If the transformer reduces the voltage by half it will exactly the double the current when there is a load on the transformer. Is this correct?

No. Always.

 

[yungman]

Then why do you say that the current capability and voltage ratio of a transformer is kind of independent?

The original question is very confusing. If all asked was whether a transformer that STEP DOWN from 110V to 11V, then if the secondary is drawing 1A, then the primary is drawing 0.1A since the step down ratio is 10:1.

I really got thrown off by "the transformer increase the current"! I was thinking you refer to current capability of the transformer, that has nothing to do with the stepping down ratio, it has to do with what the transformer rated.

Usually people refer the transformer is a step down transformer, then the input current to output current ratio increase like the example I gave on top.

 

[sophiecentaur]

You can't just "increase the current" into a load without changing the volts or the resistance of the load. Some of this thread is confusing Cause and Effect.

The turns ratio of the transformer determines the ratio of input volts to output volts.
Vs = N Vp (the first step in the argument).

The output volts will cause current to flow in the load (Secondary Current = Secondary Volts / Load resistance)
The secondary current then determines the primary current.
Ip = N Is

In a real transformer, in which there is finite resistance in the windings and in which the magnetic flux is not linked 100%, Ip may be a bit more and Vs may be a bit less than the above formulae suggest, particularly under load.
So there are two answers to the question. In fact, many off-the-shelf transformers behave in a pretty 'ideal' way.

 

[Phrak]

Without reading all the threads, in addition to vk6kro and sophiecentaur have said:-

The idea transformer has the relationship

I_pV_p = I_sV_s:

Power delivered into the primary is the power out of the secondary, where the V's and I's are functions of time.

Real transformers are not perfect:

1)There is a "magnetization current" in the primary when there is no load on the secondary. This is fairly constant, independent of load.
2) There are series resistive losses in both the primary and secondary, as someone mentioned.
3) There are core losses. It takes energy to magnetize and demagnetize the core that shows up as heat.
4) Mutual inductance. Two inductors close to each other may make a transformer, but a poor one. Mutual inductance says how much of the magnetic flux passing through the primary also passes through the secondary. Ideal transformers have a mutual inductance of one.
5) Transformers radiate energy and electromagnetically couple to surrounding circuits and conductive material.
6) There may be 6, 7, or more I can't think of, right off.

...Umm. Here's one I forgot. Transformer windings will have mutual capacitance. This changes how the voltages and currents are related as well.

 

[russ_watters]

I also thought the question (and some of the answers) was cumbersome. My take on it is that you can't have a voltage drop across a transformer without a current, so there is no realistic case where they can be independent of each other except due to losses.

The only realistic situation I can think of where they'd be different is if you had an energized primary and open secondary. Then you could have a voltage without a current at the secondary. But still, that's just a 100% loss situation.

 

[yungman]

I also thought the question (and some of the answers) was cumbersome. My take on it is that you can't have a voltage drop across a transformer without a current, so there is no realistic case where they can be independent of each other except due to losses.

The only realistic situation I can think of where they'd be different is if you had an energized primary and open secondary. Then you could have a voltage without a current at the secondary. But still, that's just a 100% loss situation.

I got thrown off too, I think he is just refer to step down ratio and primary to secondary current ratio only. OP need to clarify his question.

 

[Antiphon]

All of the confusion is stemming from treating the transformer as a source or load. In fact it is neither.

A resistor is a load and defines a ratio of voltage and current at it's terminals. No matter what source it's connected to, this ratio will never change. If the voltage goes down, the current also goes down.

A source is different. Whether a voltage or current source, it has no such defined ratio. It depends on the load that is connected.

So far so good. Now consider the ideal 2:1 step-down transformer. If you connect it to a 100 volt source, the secondary voltage is 50 volts. The primary and secondary current are unknown and could be anything. Once you put a load on the secondary, then that load defines the current, not the transformer.

What the transformer enforces at all times is that the OUTPUT voltage will be half of the INPUT voltage and current will will observe the inverse relationship.

The original post asked whether transformers increase current when they decrease voltage. The answer is yes and the question is not ambiguous or confusing unless you are considering the secondary current and the secondary voltage only. As I have shown, that ratio is completely independent of the transformer and depends only on the load.

The confused answers were not considering the primary-to-secondary conversion action of thranformers which is their only defined behavior. Bringing is copper losses and ratings only increased the confusion and were not referenced either implicitly or explicitly by the OP.

 

[Phrak]

All of the confusion is stemming from treating the transformer as a source or load. In fact it is neither.

A transformer primary with an open secondary is an inductive load with series resistance to good approximation in most cases.

A transformer primary with an loaded secondary is an inductive load with series resistance and parallel load impedence to good approximation in most cases.

 

[sophiecentaur]

All of the confusion is stemming from treating the transformer as a source or load. In fact it is neither.

. . . . . .

The original post asked whether transformers increase current when they decrease voltage. The answer is yes and the question is not ambiguous or confusing unless you are considering the secondary current and the secondary voltage only. As I have shown, that ratio is completely independent of the transformer and depends only on the load.

OP.

The problem with the original question was that it seems to imply that halving the secondary voltage actually doubles the secondary current - which is not true at all - in fact, if the load is the same, the current will also be halved.

The simplest equation for a transformer,
V₁I₁ = V₂I₂
was, I'm sure, the only thing that the OP refers to - nothing fancy about the real behaviour of a transformer with inductance and resistances involved. As could be expected on a PF thread, contributors have introduced loads of sophisticated factors which, while true, valid and interesting, can only confuse someone who just wants to clear up the basics of transformer operation.

That basic equation just tells you that, if the voltage is stepped down, the input current is proportionally less than the secondary current - that is the 'causal' line of events. The transformer doesn't, in some way, 'force' more current out of the secondary. The secondary current will just depend on the secondary volts and the load.

This is analogous to a lever used as a velocity multiplier against a frictional load (Short end on the effort side) The load end of the lever will exert a smaller force than the effort and could move further than the effort end - but only if the load friction is small enough to allow it. A rather clunky equivalent, I admit but I never really get on with analogies.

 

[vk6kro]

The equation
V1*I1 = V2*I2
just says that "power in = power out".
This applies directly if the transformer losses are negligible, as they should be.

So, suppose you have a transformer with two secondaries. 120 volt primary and 12 volt and 6 volt secondaries.

Suppose you put a 12 watt load on the 12 volt secondary.
There will be a current of 1 amp drawn from the 12 V secondary and there will be an extra 12 watts entering the transformer at the primary. 12 watts at 120 volts is 0.1 amps.

Now put a 12 watt load on the 6 volt secondary.
There will be a current of 2 amps flowing.
At the primary of the transformer, there will be 12 watts entering the transformer (or an extra 12 watts if the first load is still in place). So the primary current will be an extra 0.1 amps.

Note that the current and voltage are inversely related so that the power stays the same in each case.
Also note that the result is the same whether the transformer is at 5 % of its rated capability or 90%.

 

[srikanthhp]

How do i make current reduction

I have a transformer with a rating as input 230V Ac and output as 12V AC with 6V AC and a GND inbetween 12V and 6V.i.e., 12,6,0,-6,12 is what the connections say.The output is rated as 3A .I am having a stepper motor which has a rating of 1.5A ,so how to reduce the current drawn from the transformer.Also can I connect both 6A as well as 12 A from the same transformer?Also can I take multiple circuits to act as load so the current in each circuit follows Kirchoffs law and divides the 3A?

 

[Windadct]

Since we tend to think of the transformer as a source - at least a voltage source, the language should then be "AVAILABLE CURRENT" - as in the OP. If the secondary is open circuited ( no load applied) the secondary voltage is related to the primary almost exactly the winding ratio ( 2:1, 100:1 etc) - but the current is not. Once you apply a load to the secondary creating some secondary current - THEN the current in the primary is also proportional by the same ratio. Some of the best tests in "thinking" about these cases is the boundary conditions - the open secondary case should help.

The half the voltage ...therefore twice the current thinking is misleading - the transformers winding ratio does not FORCE the current.

Of course - this is a way of thinking, based on ideal models, the more accurate or different cases ( like a current transformer) - require different thinking, this is a way to simplify the whole process. Is it 100% accurate? no - but in most cases it tells us all we need to know.

 

[sophiecentaur]

This is a very common problem when starting on electrical matters.
You need to think of these things in the right order. Th transformer ratio will give you the secondary volts, if you know the mains supply volts. The secondary volts from the transformer and the resistance of the load will give the secondary current. The primary current is then scaled ('the other way up') according to the transformer ratio.

 

[Prabhakaran R]

After reading all, I am not getting the answer I require..
1) Consider a motor is connected at the secondary of transformer and its running at a certain speed. Now if the primary voltage is reduced, definitely motor will take more current to maintain its speed. So in this case, current increases when I reduce the primary voltage.
2)Now, Consider a resistive load is connected at the transformer seconday.If I reduce the primary voltage,the current also will reduce but incase of motor it increases. So what causes to increase the current is whether the load is resistive or inductive am I right??
If its truly depends on load whether it is resistive or inductive, The transformer itself a inductive similar to motor.So according to this,eventhough I connect resistive load, if I reduce the primary voltage, the current should try to increase upto some limit and then start reducing?? Please clarify me clearly..

 

[sophiecentaur]

1) Consider a motor

2)Now, Consider a resistive load is connected at the transformer seconday

These are not equivalent situations so any differences do not constitute a violation of principle. A resistive load on a transformer will pass the current, appropriate to the applied voltage. Lower secondary volts will give lower current etc.
A motor is not a simple resistor and the current / voltage characteristic will depend on the design of the motor. The motor you refer to would be 'unusual' if it maintains its speed, independent of the supply volts; it would need some speed regulation in its circuit and it would definitely be a non-ohmic load on the transformer. The existence of a transformer in the circuit is not particularly relevant to how it performs - except for the obvious difference in primary (supply) volts and secondary volts.

 

[CWatters]

Here is a general rule that works for the majority of simple transformer circuits...

When analysing voltages start at the power source and work towards the load.
When analysing currents start at the load and work back towards the power source.

Example:

Suppose your power supply is 240V and you motor is 120V. Let's assume the motor is lightly loaded (delivering just 20W) and is 100% efficient. You will need a step down transformer 240V to 120V and we will assume that's also 100% efficient.

Voltages first...

You start with a 240V source and go through a 2:1 step down transformer to produce 120V for the load/motor.

Current...

Start with the load.. The motor is delivering 20W and since it's 100% efficient it draws 20W from the transformer. 20W at 120V means the secondary current is 20W/120V = 0.167A. It's a step down transformer so the primary current is half of the secondary current. This means the primary current is 0.167/2 = 0.084A. This means the current delivered by the 240V source is 0.084A.

Changes..

If you change anything (for example the winding ratio of the transformer or the load on the motor) repeat the process. Work "forwards" to recalculate the voltages and "backwards" to recalculate the currents.

 

[CWatters]

PS: This will work for most loads (eg motors, lamps, heaters, inductive, resistive etc). Where it may not work is when you have current limiting devices in the circuit. For example some sort of current limit in the power supply that reduces the voltage when there is an overload.

PPS: In the real world transformers have winding resistance so when you change the load current the voltages can change, but perhaps life's complicated enough already :-)

 

[russ_watters]

The motor you refer to would be 'unusual' if it maintains its speed, independent of the supply volts; it would need some speed regulation in its circuit ...

3-Phase induction motors do indeed behave this way, with or without control. RPM is a function of frequency and Volts and amps vary opposite each other.

This can cause an issue at start-up, when extra torque/power is needed and the amperage spikes, particularly if you don't have something to soften the start. Variable frequency drives, which do control motor speed, make voltage proportional to frequency and hold amperage constant for constant torque.

 

[sophiecentaur]

3-Phase induction motors do indeed behave this way, with or without control. RPM is a function of frequency and Volts and amps vary opposite each other.

Whoops! Of course, you are right. I was only thinking of an AC/CD motor.