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If a step-up transformer increases voltage, then how current can be decreased ?

there is something very-much paradoxical !!!!

If a step-up transformer increases voltage, then how current can be decreased ?

(I'm trying to ask that, in case-of battery, if I take a battery of lower voltage (say 10 Volt), and another battery of higher-voltage (say 50 volt), and use them in 2 separate circuits (each with same path-resistance , say 2 Ohm ) ; then , in the circuit that got stronger-battery (50 V), will give much more current.

Now, instead of batteries; we're taking 2 AC sources: an AC source of 10 V , and another similar AC-source of 10 V magnified to 50 V through a step-up transformer.

Since a step-up transformer increases voltage and decreases current; then , that 50 V AC source should give LESSER CURRENT than the 10 V (according to conservation of energy).

But common sense is telling, if there is a potential difference of 50V, then we SHOULD get a MUCH GREATER CURRENT THROUGH OUTPUT(Load) circuit, than the 10V source; since we know from Ohm'law (and even non-ohmic conductors with any positive resistance), that, if we apply greater voltage, we'll obtain greater current.

Then, among these 2 thoughts (assumptions) , which one is correct? and why?

& if the first assumption is correct ( "the 50 V transformer gives 1/5 times lesser flow than the 10V original source", then why we're telling it 50V at all; and not saying 10/5 or 2 volt? )

And if it is really works just like a true 50V source and increase current, then how it obey at all the law of Conservation of energy?

See also: How do Step Up Transformer work? on this website (https://electronics.stackexchange.com/questions/11004/how-do-step-up-transformers-work)

and

On that page, an answer , explaining how each loop work as a battery. https://electronics.stackexchange.com/questions/11004/how-do-step-up-transformers-work/11006#11006

However I'm not a physics or math person. So please explain that way.

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    \$\begingroup\$ Pin = Pout. If you change to a step-up transformer and keep the same load, then the input current will go up as well. \$\endgroup\$ – Tom Carpenter Apr 23 '16 at 11:56
  • \$\begingroup\$ so you are are saying that a step up transformer should create energy from nothing? you have solved our energy crisis! as Tom above say power in = power out. p = vi so voutiout = viniin, iout = (vin/vout)*iin. Actually you probably lose a little in heat as part of the conversion, nothing is free. \$\endgroup\$ – old_timer Apr 23 '16 at 12:06
  • \$\begingroup\$ Those resistors you have on the primary side will make it so the voltage across the primary winding is NOT 10V! \$\endgroup\$ – immibis Sep 23 '16 at 1:53
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A transformer essentially converts between voltage and current using a magnetic field. Because it is a conversion, then if the process is 100% efficient, then the output power and input power must be equal:

$$P_{in} = P_{out}$$

If they are not equal, then either you are losing energy in the transformer (inefficiencies), or gaining energy (perpetual motion anyone?!). The former can happen, the latter can't.

So based on this, what can we say about the voltage and current? Well, we know that:

$$P = I V$$

So: $$I_{in}V_{in} = I_{out}V_{out}$$

Lets say that you have a step up transformer with 10 turns on the primary and 50 turns on the secondary. This means you have a turns ratio of:

$$n = \frac{50}{10} = 5$$

So that means that the voltage will be increased by a factor of 5 (\$V_{out} = 5\times V_{in}\$). So what happens to the current?

$$I_{in}V_{in} = 5V_{in}\times(I_{out})$$

In order for both sides of that to stay equal (can't get energy from nothing!), then the current must be divided by 5. Basically you can say that:

$$I_{out} = \frac{1}{n}I_{in}\space\space\space\space\space\space\space\space\space V_{out} = nV_{in}$$


So what happens if you have a fixed load and change the number of turns? Lets do an example. We shall say that the input voltage is \$10\mathrm{V}\$, the transformer initially steps up by a factor of \$n=1\$, and then later by a factor of \$n=5\$. In both cases the output load is a \$2\Omega\$ resistor.

In the first case, your calculations are correct.

$$V_{out} = n\times V_{in} = 10\mathrm{V}$$ $$I_{out} = \frac{V_{out}}{R_L} = \frac{10}{2} = 5\mathrm{A}$$ $$I_{in} = n\times I_{out} = 5\mathrm{A}$$

Now lets go for \$n=5\$.

$$V_{out} = n\times V_{in} = 5\times 10 = 50\mathrm{V}$$ $$I_{out} = \frac{V_{out}}{R_L} = \frac{50}{2} = 25\mathrm{A}$$

Great, these match what you are saying. But this is where everything changes. We do the last step of the calculation:

$$I_{in} = n\times I_{out} = 5\times 25 = 125\mathrm{A}$$

Ahh, there we go. Notice that the input current goes up, quite considerably. This makes the scales balance so to speak - power in goes up in order to cope with the large power requirements of the load.

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    \$\begingroup\$ Why the downvote? \$\endgroup\$ – Tom Carpenter Apr 23 '16 at 12:50
  • \$\begingroup\$ Oh. that means , on the output-circuit of transformer , when the current is allowed, it also induces increase-of current in input-circuit of the transformer? i.e. in my diagram, I've considered the input circuit with 2 Ohm, 10 V, 5 Amp irrespective of output circuit switched on or off. If the OUTPUT CIRCUIT TURNED ON, then the flow on input circuit will AFFECTED, INCREASED, & become 125 A? Have I got it correctly? \$\endgroup\$ – Always Confused Apr 23 '16 at 16:02
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    \$\begingroup\$ @AlwaysConfused correct. The load on the output will directly affect the load on the input. In fact you can say that the output is "referred" to the input: \$R_{in} = \frac{R_{out}}{n^2}\$. \$\endgroup\$ – Tom Carpenter Apr 23 '16 at 16:48
  • \$\begingroup\$ @AlwaysConfused also, in my first comment I wasn't talking to you. Somebody downvoted the answer and I was asking them why if they happened to look back. \$\endgroup\$ – Tom Carpenter Apr 23 '16 at 16:49
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    \$\begingroup\$ The equations are just as valid for a step down. \$n\$ is now going to be <1. E.g. 10:1 will be \$n=N_s/N_p=1/10 = 0.1\$. \$\endgroup\$ – Tom Carpenter Apr 25 '16 at 11:41
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A transformer can not create power so a step-up in a sense both increases the current and reduces it.

If we have 10V ac supply and connect a 10 ohm resistor across it we will have 1 amp flowing in the resistor. If we now disconnect and replace it with a 2:1 step-up transformer connecting the same 10 ohm resistor across the secondary then the resistor will have 20 volts across it so will have 2 amps flowing in the resistor. Thus the current in the resistor has increased as you have pointed out.

However, this is not the sense in which we mean a step-up transformer reduces the current. If we consider the power in the resistor in our second case we have 2 amps and 20 volts making a total power of 40 watts. We therefore need at least 40 watts to be flowing into the primary. This means the current into the primary has to be at least 4 amps because we only have a 10 volt supply. In practice we will have slightly more than this because no transformer is 100% efficient, there are conduction losses in the windings and some power is needed to magnetise the core but the current may only be slightly higher than this as efficiencies in excess of 90% are easily achievable.

When we say a step-up transformer reduces the current we mean we have less current in the secondary than we do in the primary.

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Let's assume the transformer is ideal(i.e no power losses). A transformer conserves power, that is, if you're consuming 12 W at the secondary side, the same amount of power is drawn from the power source of the primary side(ex mains).

For your example: The no-load output voltage is 50 V. If you connect a load of, say 100 ohm, a 0.5 A (RMS) current would flow in the secondary side, while 2.5 A (RMS) would be drawn from the 10V AC source.

What you must understand is that current drawn from the AC source depends on the current at the secondary side.

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Here is a more complete analysis, based on my discussions with Tom Carpenter above (please see our comments under his post).


Let us establish some terminology first:

  1. \$ V \$ is the voltage of the AC source.
  2. \$ R_{1} \$ is the resistance of the wiring of the primary circuit.
  3. \$ R_{2} \$ is the resistance of the wiring of the secondary circuit.
  4. \$ n_{1} \$ is the number of turns in the primary coil.
  5. \$ n_{2} \$ is the number of turns in the secondary coil.
  6. \$ V_{1} \$ is the back e.m.f. induced in the primary coil.
  7. \$ V_{2} \$ is the e.m.f. induced in the secondary coil.
  8. \$ V_{3} \$ is the voltage drop across the wiring of the primary circuit due to resistance.
  9. \$ I_{1} \$ is the current in the primary circuit.
  10. \$ I_{2} \$ is the current in the secondary circuit.

The first five quantities are assumed to be already known, while the last five quantities are to be expressed in terms of the first five.


Now, form the following five equations:

  1. \$ V = V_{1} + V_{3} \$, by Kirchoff’s Second Law.
  2. \$ V_{3} = I_{1} R_{1} \$, by Ohm’s Law.
  3. \$ V_{2} = I_{2} R_{2} \$, by Ohm’s Law.
  4. \$ I_{1} V_{1} = I_{2} V_{2} \$, by the Law of Conservation of Energy.
  5. \$ \dfrac{V_{1}}{n_{1}} = \dfrac{V_{2}}{n_{2}} \$, by Faraday’s Law of Induction.

Plugging Equation (2) into Equation (1) yields $$ V = V_{1} + I_{1} R_{1}. $$ From Equation (5), we have \$ V_{1} = \dfrac{n_{1}}{n_{2}} V_{2} \$, so $$ V = \frac{n_{1}}{n_{2}} V_{2} + I_{1} R_{1}. $$ Using Equation (3), we find that $$ V = \frac{n_{1}}{n_{2}} I_{2} R_{2} + I_{1} R_{1}. $$ From Equations (4) and (5), we have \$ I_{2} = \dfrac{n_{1}}{n_{2}} I_{1} \$, so $$ V = \left( \frac{n_{1}}{n_{2}} \right)^{2} I_{1} R_{2} + I_{1} R_{1} = I_{1} \left[ R_{1} + \left( \frac{n_{1}}{n_{2}} \right)^{2} R_{2} \right]. $$ Solving for \$ I_{1} \$, we therefore obtain $$ I_{1} = \frac{V}{R_{1} + \left( \dfrac{n_{1}}{n_{2}} \right)^{2} R_{2}}. $$ Consequently, \begin{align} I_{2} & = \frac{\left( \dfrac{n_{1}}{n_{2}} \right) V} {R_{1} + \left( \dfrac{n_{1}}{n_{2}} \right)^{2} R_{2}}, \\ V_{3} & = \frac{V R_{1}}{R_{1} + \left( \dfrac{n_{1}}{n_{2}} \right)^{2} R_{2}}, \\ V_{1} & = \frac{\left( \dfrac{n_{1}}{n_{2}} \right)^{2} V R_{2}} {R_{1} + \left( \dfrac{n_{1}}{n_{2}} \right)^{2} R_{2}}, \\ V_{2} & = \frac{\left( \dfrac{n_{1}}{n_{2}} \right) V R_{2}} {R_{1} + \left( \dfrac{n_{1}}{n_{2}} \right)^{2} R_{2}}. \end{align}


Conclusion: Ohm’s Law is in harmony with the Law of Conservation of Energy.

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  • \$\begingroup\$ @AlwaysConfused: Hi. I was wondering if you’ve managed to understand my solution to your problem. \$\endgroup\$ – Berrick Caleb Fillmore Sep 22 '16 at 22:14
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To give an insight why a transformer works this way, understand that the strength of a magnetic field (formally, the magneto-motive force or MMF) is measured in Ampere-Turns.

So you apply an AC voltage (\$ V \$) to the transformer’s primary (of \$ N \$ turns), and that drives a specific current (\$ I_{0} \$) through the primary inductance, and that current creates a magnetic field of \$ N I \$ ampere-turns.

So far, the secondary is open-circuit and we take no power out of it, which is why I’ve labelled this current as \$ I_{0} \$. It generates the magnetic field, so it’s called the “magnetisation current”.

Now, the current \$ I_{0} \$ can be calculated from the primary inductance \$ L \$, the driving voltage and the AC frequency by standard AC formulae. You’ll find this if you look, but the important point is, this is all wasted power, so you want \$ L \$ (and therefore \$ N \$, since \$ L = N^{2} A_{L} \$) to be large enough to keep wasted power down to a few percent. (Here, \$ A_{L} \$ is the ‘specific inductance’, which is a property of the transformer core).


Now, what happens if we draw a current \$ I_{2} \$ from the secondary, with \$ N_{2} \$ turns? This current also creates a magnetic field, of \$ - N_{2} I_{2} \$ ampere-turns, i.e., in the opposite sense to the field created by the primary. (because \$ I_{2} \$ is drawn from the secondary instead of fed into it.)

This decrease in the magnetic field reduces the ability of the primary to block primary current flowing (i.e., its impedance), so primary current increases until the MMF is back to the original \$ N I_{0} \$ ampere-turns. (This is for a perfect transformer. A real transformer doesn’t work quite as well as you have to consider the ‘leakage inductance’, but ignore that for now.)

So the primary current is now \$ N I_{0} + N I_{1} \$, where \$ N I_{1} \$ generates MMF to exactly cancel the secondary current’s MMF, so $$ N I_{1} = N_{2} I_{2} \qquad \text{or} \qquad I_{1} = \left( \frac{N_{2}}{N} \right) I_{2}. $$ In other words, for a step-up transformer where \$ N_{2} > N \$, the primary current must increase to generate the same MMF.

So the secondary current is determined by the secondary voltage and the load, and the primary current is determined by the secondary current (plus \$ I_{0} \$ the ‘magnetisation current’).

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