# Why does current decrease when voltage is increased? (transformers)

Why is the current in the secondary coil less than in the primary when the voltage is greater in that coil as compared to primary coil? (for transformers)

Well I know that energy should be conserved but listen this is not the cause! There would be some motion, forces or fields included in reason that finally results in high voltage and low current.

And i also think that solution will obey ohms law too since I am not talking about non ohmic conductors. If you think it doesn't then give explanation for why it obeys when you use a battery and a metal wire and why it does not apply for transformers.

• what about the POWER!!! ;D Dec 23, 2021 at 10:09
• Directly saying due to conservation of energy or P is directly proportional to V and I is not a reason for why it happens so.
– user316791
Dec 23, 2021 at 10:12
• @user6760 What if i ask you why V=IR in case of using a 4V battery jn simple circuit? Would the reason be because of conservation of energy or power? : )
– user316791
Dec 23, 2021 at 10:14
• Ohm's law is a description of how resistors work. It doesn't apply to any devices except resistors. It doesn't apply to capacitors, inductors, transistors, diodes, SCRs, vacuum tubes, ..., or transformers. Dec 23, 2021 at 17:21
• The total current through the transformer doesn’t change. It simply is divided among different number of coil wounds. The magnetic field doesn’t care how many turns the wire does. The total current sum in all the wounds is the same in the primary and secondary coils. Dec 26, 2021 at 10:25

If the output from the secondary of a transformer is connected to a fixed load (such as a resistor), an increased voltage will produce an increased current. This will require an increase in the current in the primary (in phase with the input voltage to match the output power).

P = VI

I = P/V

So if we keep the load constant then it means that power to be delivered to the load from secondary is also constant.

So in this case the load which is constant will draw more current if the secondary voltage is decreased. And load will draw less current if the voltage is increased. It's the load which determines the amount of current flow through it.

• Power is not delivered! Electric fields and magnetic fields are delivered. Power comes later. Don't be too fast. Explain for electromagnetism in this Dec 25, 2021 at 15:26

To answer your question it is necessary to explain how a transformer works. Also you have to understand the complex notation for currents and voltage. I am not going to explain it, if you don't understand it, please check some physics textbooks.

Both coils of the transformer are wound around the same core. Let us say that the primary goes around the core $$N_1$$ times and the secondary $$N_2$$ times.

Depending on the size of the core and the exact nature of the ferromagnetic substance in it the magnetic flux in the core caused by a current $$I$$ in a single loop will be $$\Phi =\ell I$$. Consider a current $$I_1$$ in the primary and a current $$I_2$$ in the secondary. For convenience I am counting the relative directions of the currents in opposite ways so the total flux in the core is

$$\Phi_{core}=\ell (N_1 I_1-N_2 I_2)$$

The total flux in the primary is $$\Phi_1 = N_1\Phi_{core}$$ because it wounds $$N_1$$ times, while the in the secondary it is $$\Phi_2 =N_2 \Phi_{core}$$

The calculations would be much more complicated if one takes into account Ohmic losses in the coils. Therefore I will suppose that the coils are perfect conductors, and ignore Ohmic losses in them. This of course is not exactly true, but it better be a good approximation, otherwise your transformer will cost you a lot of money in charges from the Electricity Company...

In the absence of Ohmic losses inside the transformer, the voltage difference between the endpoints of the primary and secondary are $$-d\Phi_1/ dt$$ and $$-d\Phi_2/ dt$$ respectively. In complex notation the time derivative, for an alternating current of frequency $$\nu$$ (for instance 50 periods per second) is obtained by multiplying by the complex number $$-i\omega$$ where $$\omega=2\pi \nu$$. To simplify notation I will call $$L$$ the quantity $$2\pi \nu \ell$$ but remember, here $$L$$ already takes into account $$\omega$$ and thus my notations differ from the usual ones.

With all these preliminaries the voltage at the primary is $$iLN_1(N_1 I_1-N_2 I_2)$$. I also assume your Electricity Company has enough power to guarantee that this voltage is indeed the promised one, namely $$V$$, be it 110V or 220V, whatever amount of electricity you use (assuming there are no losses in the wiring inside your apartment, which is not exactly true but better be also a good approximation.....)

So at the primary we have $$V=iLN_1(N_1 I_1-N_2 I_2)$$ and therefore at the secondary the voltage is $$iLN_2(N_1 I_1-N_2 I_2)=(N_2/N_1) V$$ If $$N_2$$ is larger than $$N_1$$ you have a voltage raising transformer.

Now what are you doing to do with this output voltage ? If you run an electric motor, things will be too complicated. Just assume you run it through a pure Ohmic resistor, for instance a light bulb of resistance $$R$$.

Then the current $$I_2$$ in the secondary circuit will be
$$I_2=(N_2/N_1) V/R$$

Now we can calculate $$I_1$$. We have $$iLN_2(N_1 I_1-N_2 I_2)=R I_2$$ Thus $$iLN_2N_1 I_1=(iLN_2^2+R )I_2 \to I_1=(N_2/N_1-iR/(LN_1N_2)\ )I_2$$

The ratio $$I_1/I_2$$ is the complex number $$N_2/N_1-iR/(LN_1N_2)$$ ! In normal utilisation context, $$R$$ is small compared to $$LN_2^2$$ (remember that in my notations $$\omega$$ is already counted into the definition of $$L$$ which is not the usual notation). So the imaginary part is small, and the ratio $$I_1/I_2$$ is close to the real number $$N_2/N_1$$.

In the case of a voltage raising transformer, the current in the primary is larger than in the secondary by a complex number close to the real number $$N_2/N_1$$, which can be expressed as

In the case of a voltage raising transformer, the current in the secondary is smaller than in the primary by a complex number close to the real number $$N_1/N_2$$, in normal utilisation.

But what happens if $$R$$ is large ? Than the imaginary part of the ratio $$I_1/I_2$$ becomes larger than the real part ! But it does not mean that $$I_1$$ becomes larger, it means $$I_2$$ becomes smaller. In fact $$I_1=(\ (N_2/N_1)^2/R-i/(LN_1^2)\ )V$$ If there is nothing plugged into the secondary, it amounts to an infinite $$R$$. $$I_2$$ is zero, of course, but not $$I_1$$ (unless you also unplug the primary). But then this current is purely imaginary.

What ! When nothing is plugged in the secondary, there still is a current in the secondary ? And the company is charging me for it ?

No. An imaginary current does not produce power. You are only charged for power, not current. The meter of the Electrical Company only measures the real part of the current, the power producing one, not the imaginary one. It is built that way on purpose.

It you consider the power, indeed in the resistor, you have voltage $$(N_2/N_1)V$$ and current $$I_2$$. The real part of $$I_1$$, the only power producing one is $$(N_2/N_1)I_2$$ and delivered at potential $$V$$. So the power is indeed the same.

But you are right. Just the argument of power would not allow you to compute the imaginary part of $$I_1$$, which is present even with nothing plugged into the secondary.

But nothing is perfect, and there are Ohmic losses everywhere, the current $$I_1$$ is never purely imaginary when the secondary is unplugged, and thus the Electricity Company will measure its real part and charge it to you. So to save money and help our planet, do unplug the primary when you do not use the secondary.

• Thus $iLN2N1I1=(iLN22+R)I2→I1=(N2/N1−iR/(LN1N2) )I2$ On the most right hand side, i should be below $R$. Check it
– user316791
Feb 2, 2022 at 4:50
• @ArsenalCreation $i^2=-1$ So indeed $+R/i$ is the same thing as $-iR$. The usual convention, which I have followed, is to avoid $i$ at the denominator, always bring it to the numerator by changing the global sign. Feb 2, 2022 at 4:59
• "If there is nothing plugged into the secondary, it amounts to an infinite R" So you are saying that vaccum has infinite resistance. It is not true. How do you came up with term "inaginary current "?
– user316791
Feb 2, 2022 at 5:03
• What do you mean, it is not true ? For all practical purposes, if you don't plug anything, there is no current flowing. There might be some leaks, the plastic insulators are never perfectly insulating so you do have a non-infinite resistance, but that goes through matter, not vacuum. Current through air would cause a visible spark. Feb 2, 2022 at 5:10
• @ArsenalCreation "Imaginary" current means in quadrature of phase with voltage. "real" current means in phase with voltage. It can all be done with sines and cosines, but it is a pain in the neck. Complex notation for currents when self-inductance and also capacitors enter the problem are standard notation. Feb 2, 2022 at 5:14

It’s the other way around: increasing current cancels some of the B field in the coils, which reduces the induced voltage.

The voltage induced in the coil is due to the (time rate of change of) the total magnetic field. Both the current in the primary and secondary currents contribute to that field. By Lenz's law, the secondary current reduces the magnetic field. The more secondary current -> the less total magnetic field -> the less change in the magnetic field -> the less voltage induced on the secondary.

In the limit of no current in the secondary, you get the maximum voltage.

In the limit of no resistance in the secondary (shorted), you get the maximum possible current, which is just enough to cancel the complete magnetic field, which in turn results in no voltage across the output: That's what's needed if the output is shorted.

You can make this all mathematical if desired, but the qualitative form is important to understand.

• Can you explain it in detail if it is correct?
– user316791
Jan 30, 2022 at 14:01
• Updated the answer to have more detail. Jan 31, 2022 at 16:15
• If i assume that changes(current and voltage) taking place inside secondary coil are quick or almost instantaneous then, would not the current change in secondary coil with respect to voltage across it. So even if you produce a current in secondary coil that reduces the initial voltage across it, the current will also change.
– user316791
Feb 1, 2022 at 8:45
• Transformers are generally used in AC (sine and cosine) circuits. You can also analyze for more complex waveforms, but it’s more complex. Note that if you “produce” a current, that’s the current. Period. Then only the voltage can vary in response. Feb 1, 2022 at 23:11