# Transformers: relation between their current, voltage and resistance

My current understanding

Transformers are used to step up and down voltage keeping power constant. Hence, for example, if I step up some voltage, the current will decrease in the secondary circuit.

The issue I have

I am not able to digest the fact that on applying more voltage across a circuit, current decreases. To me it seems that the extra potential difference should make the electrons zoom faster.

My guess

It occurred to me that maybe, for example during step up, due to the fact that there are more coils in the secondary circuit, the resistance of the secondary circuit increases and this increase in resistance reduces the current, although there is a higher voltage.

But then I thought that if we use another material in the secondary coil, that has lesser resistivity, such that even on making more number of turns, it has a resistance equal to that if the primary coil, then what happens?

The analogy I want to use

10 joules of energy can apply 10N for 1m or 5N got 2m. But it does the same amount of work because the first object is obviously heavier.

I would really appreciate it if someone could:

1. Clarify how there is increase in voltage but decrease in current in secondary coil.

2. Explain why my guess is correct or wrong

3. Relate my analogy to the problem.

4. Correct anything wrong in my understanding.

Thank you

Edit:

There is one more similar question on the site, but it has a mathy answer. I'm looking for something more intuitive. (However, please include the relevant formulas.)

• That's it! You got answers!! All the best :D – user36790 Feb 15 '16 at 4:30
• Your question includes 'increase in voltage but decrease in current' in a way that suggests a component has lessened current when applied voltage is increased (i.e. negative resistance) but your 'increase in voltage' isn't associated with voltage stress on a component, rather it is associated with a circuit component change (altering the turns ratio of a transformer). It is misleading phrasing, to describe a turns ratio change as though it were negative resistance. – Whit3rd Feb 15 '16 at 8:35
• @Mahathi Your confusion stems from misconception that both sides of a transformer are "separate circuits". A transformer does not "separate" circuits, it "magnetically couples" them. So you have one large circuit, with a different primary and secondary section. Any circuit analysis ignoring this primary-secondary coupling is likely to be quite inaccurate. – ManRow Dec 25 '19 at 11:58

I am not able to digest the fact that on applying more voltage across a circuit, current decreases.

This isn't a correct picture of transformer operation.

For concreteness, assume the secondary circuit (load) is a single resistor of resistance $R_L$ and assume the primary is connected to an AC voltage source of magnitude $V_p$.

Now, if the primary voltage is increased (decreased), the current through $R_L$ increases (decreases) just as you expect it would.

However, for a step up transformer, the load resistance appears smaller to the source. So, in fact, while the voltage on the secondary is larger, the current in the primary is larger than it would be if the source were connected directly to the load.

Put another way, the source 'sees' a resistance smaller than $R_L$ and, thus, must supply more current to the primary than if the $R_L$ were connected directly to the source.

So, in fact, the increase in secondary voltage results in an increase in primary current. This is the nature of transformer action.

The analogy I want to use

The proper mechanical analogy to use is force (voltage), velocity (current) and lever (transformer).

For a step up (down) transformer, the lever is longer (shorter) on the force source side of the fulcrum.

For example, a sinusoidal force (voltage) applied to the long side of the lever, will result in a larger force (voltage) on the short side but a lower velocity (current). This is, I believe, intuitive.

What I want in the answer

I appreciate a direct approach. Assuming you do too, be aware that you are owed nothing and those that take the time to answer here do so in the expectation of trade of value for value.

• +1, thanks for the great analogy. I put 'What I want in the answer' hoping to make my question clear and easier to answer, but I guess it looked cocky or seemed like an order. I'll reword that. – Mahathi Vempati Feb 15 '16 at 3:53

Late on the question, but I thought to answer for people who are reading this later.

### "I am not able to digest the fact that on applying more voltage across a circuit, current decreases."

Your picture of what transformer really does is at least on some level wrong. I always thought that the way we are often taught that transformer "multiplies voltage and divides current or the other way round" was stupid. Obviously this cannot always hold true, for example if the secondary circuit is open so that little or no current can flow.

What actually is conserved is electromotive force, i.e. the force to induce current.

What this means is that the change of magnetic field will impose the same "force" to each loop of wire on the secondary side. This always holds true. This force persists until the current on the secondary is large enough to generate it's own magnetic field that cancels the field out.

First, if we have many loops on the secondary side, each of them will feel the same current inducing force - voltage. They are in series in a sense and the voltages add up. This is why the voltage on the secondary side can be larger.

Secondly, when the current starts to rise in the secondary wire, the magnetic field it generates is multiplied by the loop count. Obviously if we had lots of loops, a very small current will quickly generate enough magnetic field to cancel the inducing field out and there is no more force to raise the current. In other words: high voltage transformer will impose strong voltage on the secondary side, but this strength is quickly depleted.

Inversely, in a voltage lowering transformer there are few loops on the secondary side. The voltage they generate together is small, and may not easily cause strong currents. However if they are connected to a load of unusually low resistance, the current will continue to rise until the magnetic field is cancelled out, or resistance blocks further current. With only few loops the current required to to cancel the field out can be staggering.

One might want to use the analogy of a lever here. A lever where longer arm is pulled and shorter does the work the force generated is strong, but has no ability to move fast. A lever where shorter arm is pulled can generate huge speed on the other side, but carries little force.

• This is a very poor answer and incompatible with physics and electrical engineering. – Andy aka Dec 11 '18 at 22:48

I'm answering late, because this question has been referred to in a newer question with some misunderstanding of the answers here.

Hence, for example, if I step up some voltage, the current will decrease in the secondary circuit. The issue I have I am not able to digest the fact that on applying more voltage across a circuit, current decreases.

It's not the current at the secondary that drops, necessarily. It's the ratio between the primary and secondary currents.

So don't think of it as increasing the secondary voltage causes the secondary current to drop.

Think of it as, if you add turns to increase the secondary voltage, you will have to (super-linearly) increase the primary current to be able to realize the expected increase in secondary voltage.