How do transformers obey ohm's law?

[Entanglement]

If you apply 100 volts to a 10:1 transformer you will end up with 10 volts out. If you put a 10 ohm resistor on the secondary you will have one ampere passing through the resistor. That's it !' Why do I have to apply energy conversion equation to find the intensity it seems meaningless, why don't I just calculate the induced emf then find the intensity by : Induced Emf / R of secondary coil. Please guys, I'm sure that I'm wrong, so I need a good explanation for this

 

[Andrew Mason]

If you apply 100 volts to a 10:1 transformer you will end up with 10 volts out. If you put a 10 ohm resistor on the secondary you will have one ampere passing through the resistor. That's it !' Why do I have to apply energy conversion equation to find the intensity it seems meaningless, why don't I just calculate the induced emf then find the intensity by : Induced Emf / R of secondary coil. Please guys, I'm sure that I'm wrong, so I need a good explanation for this

Because the secondary is a coil, there is inductive reactance in addition to the load resistance. Ohm's law applies to resistance loads. If there is a phase angle between current and voltage, which there will be unless the there is a capacitor in there somewhere, the power is not given by P = VI. It is P = VIcosΘ where Θ is the phase angle between current and voltage. If Θ = 90° (no resistance, only inductive reactance) there is no power consumed at all.

AM

 

[Entanglement]

Because the secondary is a coil, there is inductive reactance in addition to the load resistance. Ohm's law applies to resistance loads.
AM

Can you detail that a little bit ??

 

[Philip Wood]

To a good approximation, the secondary current is given by:

$$I_s = \frac {\epsilon_s}{R_L + R_S}$$
$\epsilon_s$ = emf in secondary, $R_L$ = load resistance, $R_S$ = resistance of secondary coil.

The reactance due to self inductance of the secondary coil is, I believe, usually negligible.

The word you need is 'current', by the way, not 'intensity'.

 

[Entanglement]

I don't understand the difference between both resistances?

 

[Philip Wood]

That's easy: the secondary coil of the transformer itself has resistance. So does the 'load' you connect across the secondary coil. These two resistances behave as if they are in series. That's because of the way the emf is induced in the secondary coil.

 

[Entanglement]

How do transformers obey ohm's law??

So in a step down transformer of turns ratio 2:1 if the voltage in the primary coil is 2V the voltage in the secondary will be 1V

Then to find the current in the secondary coil we just divide 1volt by the sum of both resistances ?

 

[Philip Wood]

Yes. In many cases, though the load resistance is much greater than the coil resistance, so you can forget the coil resistance.

 

[Entanglement]

Yes. In many cases, though the load resistance is much greater than the coil resistance, so you can forget the coil resistance.

but I'm confused because this $\epsilon$_p / $\epsilon$_s = $I$_s / $I$_p

so in my previous example if the current in the primary coil is 1A according to this equation the current in the secondary coil will be 2A not 1A

that contradicts with what we said before

 

[berkeman]

So in a step down transformer of turns ratio 2:1 if the voltage in the primary coil is 2V the voltage in the secondary will be 1V

Then to find the current in the secondary coil we just divide 1volt by the sum of both resistances ?

but I'm confused because this $\epsilon$_p / $\epsilon$_s = $I$_s / $I$_p

so in my previous example if the current in the primary coil is 1A according to this equation the current in the secondary coil will be 2A not 1A

that contradicts with what we said before

Please see the "Ideal Transformer" section of the wikipedia entry to see if it clears up your confusion...

http://en.wikipedia.org/wiki/Transformer

 

[Philip Wood]

Let's do an example. Suppose an alternating p.d of 240 V is applied to the primary of a step-down transformer of with 400 turns on the primary and 15 on the secondary. Suppose also that we connect a 5 ohm resistor across the secondary (whose own resistance is much less than 5 ohm). What will be the primary current?

Step 1: Calculate the secondary voltage using the turns ratio. 9 V agreed?

Step 2: 9 V will drive a current of 1.8 A through the 5 ohm resistor. Agreed?

Step 3: If we make the approximation that no energy is dissipated in the transformer itself, then…
power in = power out, that is $V_p\ I_p = V_s\ I_s$
This gives the primary current as 0.0675 A, doesn't it?

 

[Entanglement]

Let's do an example. Suppose an alternating p.d of 240 V is applied to the primary of a step-down transformer of with 400 turns on the primary and 15 on the secondary. Suppose also that we connect a 5 ohm resistor across the secondary (whose own resistance is much less than 5 ohm). What will be the primary current?
Step 1: Calculate the secondary voltage using the turns ratio. 9 V agreed?
Step 2: 9 V will drive a current of 1.8 A through the 5 ohm resistor. Agreed?
Step 3: If we make the approximation that no energy is dissipated in the transformer itself, then…

power in = power out, that is $V_p\ I_p = V_s\ I_s$

This gives the primary current as 0.0675 A, doesn't it?

First 2 steps : I agree

Third step, why can't we just calculate the current in the primary coil from : Vp / Rp

 

[Philip Wood]

This is a good question. I'm afraid it can't be answered without going quite deeply into transformer theory; more deeply than you may need to do at your present level of study…

The answer is along these lines… There is a changing magnetic flux in the transformer core, due to the changing current in the primary and, if there is a load connected across the secondary, to the changing current in the secondary. The changing flux induces an emf in the primary, and this emf is usually much larger than the p.d., $I_p R_p$ due to the resistance of the primary coil. So, to a reasonable approximation, we can forget all about the resistance of the primary coil. The primary current is determined by the constraint that the voltage induced in it is equal to the voltage we apply across it. If we pursue this line of argument mathematically, the primary current comes out to be $\frac {n_s}{n_p} \times I_s$ to a good approximation.

At your present level of study, step 3 is probably all you need. It is perfectly valid, even though it doesn't tell you what's going on inside the transformer.

 

[Drakkith]

First 2 steps : I agree

Third step, why can't we just calculate the current in the primary coil from : Vp / Rp

Because the resistance (or impedance to be more accurate) of the primary changes depending on the power draw of the secondary. If you cut the impedance of the secondary in half, you double your current and your power draw, which means that the current and power in the primary circuit also doubles. This requires that the impedance be cut in half for the primary circuit as well.

 

[russ_watters]

First 2 steps : I agree

Third step, why can't we just calculate the current in the primary coil from : Vp / Rp

With due deference to Philip: the harder you push on electrons going through a coil - the more you try to push through, the harder they push back.

 

[Entanglement]

 

[Philip Wood]

I'm not sure how we can help without knowing where your difficulties lie.

I think Drakkith is essentially spelling out the implications of Step 3 in my post 11, though Drakkith may disagree. [The resistance (or … impedance) of the primary, referred to by Drakkith is NOT the resistance that you could measure by connecting an ohm-meter across the primary of a transformer not in use. It is an EFFECTIVE resistance or impedance. My advice would be to keep off the idea at this stage in your learning.]

I can't comment on Russ Water's post.

 

[Entanglement]

Philip wood you are so helpful I really appreciate that

I'd like you to help me with the idea of the transformer from the beginning

We'll talk about 100% ideal transformers first, because ideality makes it easy to understand the real ones after that,

I've read that the resistance of the coils of an ideal transformer is zero ohm
If we assume that there isn't a load connected to the secondary what will happen ?

 

[russ_watters]

I was trying to convey more concisely why voltage and amperage aren't linearly dependent on each other in a transformer; the more amperage you drive through the primary, the higher the impedance becomes, so it isn't a simple V=ir situation like a resistor is.

And this is why I was focusing on that:

I've read that the resistance of the coils of an ideal transformer is zero ohm
If we assume that there isn't a load connected to the secondary what will happen ?

Resistors and inductors operate completely different from each other. Zero resistance in a straight wire implies infinite current, but that isn't anywhere close to what happens in an inductor. You'll need to stop thinking of inductors as if they were resistors. A good starting point here would be to read the wiki articles on impedance, inductors and inductance.

 

[Drakkith]

I'm not sure how we can help without knowing where your difficulties lie.

I think Drakkith is essentially spelling out the implications of Step 3 in my post 11, though Drakkith may disagree. [The resistance (or … impedance) of the primary, referred to by Drakkith is NOT the resistance that you could measure by connecting an ohm-meter across the primary of a transformer not in use. It is an EFFECTIVE resistance or impedance. My advice would be to keep off the idea at this stage in your learning.]

No, I agree. The ohm-meter would measure the simple resistance portion of the impedance, which will be very low for a transformer coil.

I've read that the resistance of the coils of an ideal transformer is zero ohm
If we assume that there isn't a load connected to the secondary what will happen ?

Before you even get into the details of transformers, do you understand what impedance is? Do you know the difference between how a resistor, inductor, and capacitor affect impedance? How about how different frequency AC currents are affected by impedance? Without knowing those things I don't think you'll understand how a transformer works.

 

[Entanglement]

Before you even get into the details of transformers, do you understand what impedance is? Do you know the difference between how a resistor, inductor, and capacitor affect impedance? How about how different frequency AC currents are affected by impedance? Without knowing those things I don't think you'll understand how a transformer works.

I'm not really familiar with these concepts

 

[Philip Wood]

Responding to post 18…

The transformer primary, with negligible resistance and nothing connected across the secondary, will behave purely as a self-inductor.

There will then be a current in the primary, which is a quarter of a cycle out of phase with the alternating p.d. applied to the primary. Because of this quarter cycle phase difference, no power is taken from the alternating voltage source, over complete cycles. So when there is no power taken out from an ideal transformer, none is put in, as you'd expect.

For this answer to make sense to you, you need to understand the concept of self inductance, and how an inductor behaves in an a.c. circuit. If you don't yet have these concepts, I wouldn't try and go as deeply as this into transformer theory. Transformers are quite complicated, even when you make simplifying assumptions.

 

[Drakkith]

I'm not really familiar with these concepts

Then I would advise you to read up on those subjects, as it is impossible to understand how a transformer works without understanding what impedance is.

 

[enorbet]

We can possibly help ElmorshedyDr understand better right here, by example. Certainly everyone on this forum has some sort of sound system, possibly your PC has one. The loudspeakers are commonly listed as a simple resistance eg: 8 Ohms. This is inaccurate and the cause of many misunderstandings, since most speakers employ a voice coil, a coil of wire around a hollow cylinder which is physically attached to the vibration surface, usually the cone, and also surrounding a permanent magnet. When AC is applied to the coil it creates a magnetic field that changes polarity consistent with the frequency of the AC input and working against the fixed field of the permanent magnet it pushes and pulls the cone which moves air that we translate as sound.

If you take one of your (for this example) 8 Ohm speakers and measure it's DC resistance you will find that it is considerably less than 8 Ohms, more like 5 or 6 Ohms. This is it's DC resistance. The somewhat inaccurate 8 Ohm specification is an effective average. What is being averaged is the change of resistance of a coil to different frequencies. It might be 8 Ohms at say 1000 Hz, but perhaps 24 Ohms at 40 Hz. See? It changes because coils have resonance. Capacitors behave similarly but in an opposite direction. This is why and how coils (Inductors) and Capacitors can and are used in filtering networks and also why your sound system frequency response isn't flat "from Gravity to Light". Resistance is simple. Impedance can be a mean female dog but also an incredibly useful tool. Basically it is Static vs/ Dynamic.

 

[Philip Wood]

Re enorbet's post, it needs to be pointed out that because a loudspeaker coil moves in a magnetic field, an emf acts of which there is no counterpart in a transformer. Loudspeaker coils therefore have a 'motional impedance'; transformer coils don't.

It is, of course, true that both loudspeaker coils and transformer coils when carrying a.c. have impedances which are different from their d.c. resistances. Beyond that, I would not wish to extend the analogy.

 

[enorbet]

Re enorbet's post, it needs to be pointed out that because a loudspeaker coil moves in a magnetic field, an emf acts of which there is no counterpart in a transformer. Loudspeaker coils therefore have a 'motional impedance'; transformer coils don't.

It is, of course, true that both loudspeaker coils and transformer coils when carrying a.c. have impedances which are different from their d.c. resistances. Beyond that, I would not wish to extend the analogy.

I agree that is true in a simple air-core xformer. However when a ferromagnetic core is used there is some counterpart w/ emf effects, albeit negligible. The next step was to be to explain that the effect of a load impedance on the secondary of a xformer coil is reflected to the primary. One can't assume one knows the dynamic load measured with a static means. That only gives the ratio.

I was just trying to find some common ground to help OP relate and "visualize". Again, it's the difference between static resistance and dynamic impedance that seemed to be tripping him up.

 

[Entanglement]

This post is about 3 months olds but I like rereading your replies whenever I get confused again in the same topic,
I've grasped a pretty basic idea about self inductance, resistors, inductors and inductance reactance as I was advised in previous replies. I think that's good enough to understand transformers.
But I still don't understand why is it a must that the back emf is equal to the main supplied voltage or even approximately equal? What I know that the back emf depends on properties of the coil and the frequency of the AC, how does that make the back emf always approximately equal to the main voltage?

 

[Philip Wood]

Hello ElmorshedyDr.

Suppose a varying p.d. with instantaneous value $V_P$ is connected across the primary coil. Per coulomb passing through the primary, $V_P$ joules of work will be done. An amount $I_P\ R_P$ will be done heating the primary coil. But work will also be done per coulomb against back emfs $\varepsilon_{1, 1}$, the emf induced in the primary due to the rate of change of primary current and $\varepsilon_{1, 2}$, the emf induced in the primary due to the rate of change of secondary current.

So, using the principle of conservation of energy (per coulomb flowing),

$$V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ +\ I_P\ R_P.$$

I'm afraid that signs vary according to the sign convention used.

Now here's the punch-line… Usually $R_P$ is small enough for us to forget $I_P\ R_P$, so

$$V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ .$$

 

[Entanglement]

Hello ElmorshedyDr.
Suppose a varying p.d. with instantaneous value $V_P$ is connected across the primary coil. Per coulomb passing through the primary, $V_P$ joules of work will be done. An amount $I_P\ R_P$ will be done heating the primary coil. But work will also be done per coulomb against back emfs $\varepsilon_{1, 1}$, the emf induced in the primary due to the rate of change of primary current and $\varepsilon_{1, 2}$, the emf induced in the primary due to the rate of change of secondary current.
So, using the principle of conservation of energy (per coulomb flowing),
$$V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ +\ I_P\ R_P.$$
I'm afraid that signs vary according to the sign convention used.
Now here's the punch-line… Usually $R_P$ is small enough for us to forget $I_P\ R_P$, so
$$V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ .$$

So Vp is somehow distributed, a part of it drives a current in the primary coil as a resistor and another part drives a current in both coils as inductors ?

 

[Philip Wood]

Yes, I think you could say that, but I'd slightly reword it:

"… a part of it drives a current against the resistance of the primary coil, and another part drives the rate of change of current in both coils as inductors."

Actually I'm still not quite happy, because it's through the mutual inductance between the coils, rather than individual self inductances that an emf is induced in the primary due to change of current in the secondary.

 

[Entanglement]

Yes, I think you could say that, but I'd slightly reword it:
"… a part of it drives a current against the resistance of the primary coil, and another part drives the rate of change of current in both coils as inductors."

I don't get why you reworded it?
Why do you prefer saying "rate of change" ?

 

[Entanglement]

Hello ElmorshedyDr.

Suppose a varying p.d. with instantaneous value $V_P$ is connected across the primary coil. Per coulomb passing through the primary, $V_P$ joules of work will be done. An amount $I_P\ R_P$ will be done heating the primary coil. But work will also be done per coulomb against back emfs $\varepsilon_{1, 1}$, the emf induced in the primary due to the rate of change of primary current and $\varepsilon_{1, 2}$, the emf induced in the primary due to the rate of change of secondary current.

So, using the principle of conservation of energy (per coulomb flowing),

$$V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ +\ I_P\ R_P.$$

I'm afraid that signs vary according to the sign convention used.

Now here's the punch-line… Usually $R_P$ is small enough for us to forget $I_P\ R_P$, so

$$V_P\ \ =\ \varepsilon_{1, 1} + \varepsilon_{1, 2}\ .$$

But if the secondary coil is not connected to a closed circuit
there will be a $V_S$ but there won't be a current "to report a feedback to the primary coil" to decrease $V_P$'s share to $I_P\ R_P$
why isn't that a violation to law of conservation of energy?

 

[Philip Wood]

Re post 31
Because the back-emfs only arise when the currents are changing. Faraday's law.

 

[Philip Wood]

Re post 32
If secondary not connected then no energy is being supplied by the transformer, so the absence of feedback to the primary doesn't violate energy conservation, but is required by it!

Don't forget that voltages, such as Vs, tell us how much energy would be converted/transferred per coulomb flowing, so if there is no current (that is no charge flowing) no energy will be converted, even if there is a non-zero voltage.

 

[Entanglement]

Re post 32
If secondary not connected then no energy is being supplied by the transformer, so the absence of feedback to the primary doesn't violate energy conservation, but is required by it!

Don't forget that voltages, such as Vs, tell us how much energy would be converted/transferred per coulomb flowing, so if there is no current (that is no charge flowing) no energy will be converted, even if there is a non-zero voltage.

Why isn't there any energy supplied?

Inducing a voltage in an open circuit, as I understand, is like kicking a ball upwards against gravity then preventing it from falling by some barrier, once the barrier is removed the ball falls way back again consuming all its potential energy.
The removal of the barrier is just like closing the circuit and allowing the current to consume its voltage.
Even though the ball was prevented from falling but energy was consumed to raise it against gravity,
even though the circuit is open energy is consumed to create a potential.