Why Voltage Increases in a step up Transformer rather than the Current?

In a step up transformer, we all know voltage increases and current decreases, the answer to more voltage is because it has more windings, but why does more windings induce more voltage, can't the secondary windings generate more current but less voltage? * I've seen answers that there is less displacement current and more conduction current, some said the electrons are more congested due to more windings, and these answers were apparently wrong (said the experts from this forum).

can't the secondary windings generate more current but less voltage?

Yes, certainly. But this would be called a step down transformer. That is just a step up transformer run in reverse. It is not a matter of physics but just a matter of which sides you label as in and out. So a step up transformer by definition is one that increases the voltage and decreases the current. Take the same piece of hardware and turn it around and it becomes a step down transformer which decreases voltage and increases current.

• Yeah I know that, but I don't know why they were labelled so(voltge goes up, current goes down or vice versa), there's no way they would try one out and labelled them so without any science behind it
– user215726
Dec 14, 2018 at 12:11
• @acmilan What do you mean no science behind it? The science of how transformers work is well established. Dec 14, 2018 at 12:49
• Usually it's simply a matter of how the plugs and receptacles are connected to the various windings. The exact same transformer could be used for a 120 to 220 package as is in a 220 to 120 package, but the male and female connections are different. Dec 14, 2018 at 15:30
• @acmilan the side constructed with more windings is the “up” side and the side constructed with fewer windings is the “down” side.
– Dale
Dec 15, 2018 at 0:20
• Conservation of energy. Since V increases I must decrease.
– Dale
Dec 15, 2018 at 1:01

Transformers are made so that the same magnetic flux that goes through one coil goes through the other. Using the relation

$$\epsilon=N\frac{\text d\Phi}{\text d t}$$

for the magnitude of the induced EMF, we see that more coils induces a larger voltage. This is because each loop receives the same flux, so each loop gets an induced EMF of $$\frac{\text d\Phi}{\text d t}$$, and you can think of an entire coil as a system of loops in series.

Since the power in each coil must be equal, we have $$P_1=P_2$$, or $$I_1V_1=I_2V_2$$, so if one coil, say $$1$$, has a higher voltage, then the other coil ($$2$$) must have a higher current amplitude.

• People would say the simple P=VI formula, which makes sense but I need more elaboration and yes tge question I posted is still debugging my head
– user215726
Dec 14, 2018 at 11:49
• Shouldn't there be a negative sign for Faraday's Law? Dec 14, 2018 at 14:33
• @BobD, if $\epsilon$ is induced electromotive force, yes there should. If it is the voltage drop for ideal coil, then no, because voltage drop is minus electromotive force, to make the net field in the winding close to zero (in ideal conductor). Dec 14, 2018 at 19:50
• @JánLalinský As I understand it $ε$ in the equation for Faraday's Law is the induced voltage. Every reference I've seen shows it minus. The NCEE Fundamentals Exam (FE) for electrical engineering shows it that way. One interesting thing is in the Hyperphysics website it implies the minus sign comes from Lenz's law. Dec 14, 2018 at 20:14
• @BobD, $\epsilon$ in Faraday's law is emf, i.e. integral of electric field along some closed path. Such integral is not the actual voltage (difference of potentials), because it depends on the chosen path. However, voltage across a coil is a unique value, it does not depend on such choice. Moreover, in real coils, while emf obeys Faraday's law, voltage doesn't, because it is influenced by electric resistance of the wires. However, sometimes people say "induced voltage" when they really mean emf (more clear term). Then this "induced voltage" indeed is minus change of magnetic flux. Dec 14, 2018 at 21:04

…but why does more windings induce more voltage, can't the secondary windings generate more current but less voltage?

More windings (more turns, $$N$$) results in more voltage because of Faraday’s Law of induction given in @Aaron Stevens answer.

The question of current vs. voltage in the primary winding vs. the secondary winding of an ideal (lossless) transformer is a matter of conservation of energy, which requires that:

$$V_{primary}I_{primary}=V_{secondary}I_{secondary}$$

That is, watts into the transformer equals watts out of the transformer or, for a given time $$t$$, Joules into the transformer equals Joules out of the transformer.

Hope this helps.

Aaron Stevens answer gives you the basic answer about more windings having a higher voltage. Faraday's Law is part of EM theory, summarized in the more mathematically sophisticated Maxwell's equations. Basically, a changing current on the input side (primary) of the transformer produces a changing magnetic field (Ampere's circuit law). The changing magnetic field passes through the windings in the output (secondary) side and by Faraday's law induces an electromotive force in the secondary side which results in a voltage around the closed loop of the secondary.

The number of windings on the primary side determines the amplitude of the magnetic field and hence the size of the rate of change. Each coil/winding on the secondary side responds to the changing magnetic field. The mathematical result is that the voltage induced on the secondary side is proportional to the ratio of the secondary coils to the primary coils. As mentioned in Bob D's answer, the power must be conserved (you can't get something for nothing!) so the current must be have inversely. In short, the secondary voltage is influenced directly by the windings ratio.