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Transformer: secondary coil question 
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#1
Sep2707, 10:51 AM

P: 218

Hi
I wondered if anyone could explain as to how actually the whole process of transformation of voltage happens. I know the equations involved but can't get them together to reach a conclusion. When a.c. is passed throught the primary, a flux is induced in the core, which travels to the secondary, and induces an emf in the coil. Now emf induced is higher, I understand, because number of turns are more. But why is current lower. I know the ratio equation, but I wanted an explanation as to how actually increase in voltage leads to the decrease in current. I may sound a bit silly, but the problem is that I know what happens to emf and current separately, but am unable to connect the two things together. Mr V 


#2
Sep2707, 11:27 AM

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P: 40,730

The primary voltage across the primary inductance causes a primary current to flow, which generates the primary flux. The change in this flux through the secondary coil induces a secondary voltage, which causes a secondary current to flow through the secondary load resistance. This secondary current generates a secondary flux which opposes the primary flux, and effectively reduces the primary input impedance seen by the source that is driving the primary coil. (The input impedance effectively has the load resistance placed in parallel with the primary impedance, ratioed by the square of the turns ratio.)
Does that help? 


#3
Sep2707, 03:24 PM

P: 166

Flux = Number of turns * I So the winding which have more turns must have less current. 


#4
Sep2707, 05:39 PM

P: 1,036

Transformer: secondary coil question
A transformer cannot output more power than inputted, due to CEL (conservation of energy law). Right off the bat, it should be obvious that the secondary power, which is Isec*Vsec must equal the primary power, Ipri*Vpri, minus losses. Faraday's law, FL, relates the magnetic flux density, B, to the voltage, V, and the frequency, f. Ampere's law, AL, relates the magnetic field intensity, H, to the current, I. Also, B and H are interrelated through the permeability mu, which is analogous to Ohm's law (for magnetics). Ohm's law, OL, relate the current and voltage at the secondary with the load resistance. Most transformers are driven at their primary by an independent power source which is generally a "constant voltage" source. Current transformers can be discussed later. The voltage source magnitude, V, frequency, f, and primary turns number, Np, and the core area, A, determine the magnetic flux density, B, per FL. This magnetic flux almost completely links, or couples into the secondary winding. Since the core area is the same, as well as the frequency, only the number of turns differs, Ns. Just as FL describes the relation between Vp, f, A, and Np, it holds equally for Vs, f, A, and Ns. Again, f and A don't change, so that the ratio of volts to turns cannot change. Hence Vp/Np = Vs/Ns, since B, f, & A remain constant. When current is drawn by loading the secondary, a magnetomotive force, mmf, occurs, which tends to counter the existing core flux, which tends to reduce the voltage. But, by definition, the primary power source is a constant voltage type, which will supply whatever current needed to maintain a fixed voltage value. The primary current increases to a value needed to maintain the core flux. AL describes the relation. For a given secondary current, Is, and Ns, a magnetic field intensity, H, is given by AL. In the primary an equal and opposite H, or mmf if you will, must exist. Since H is almost equal in the primary and secondary, Np*Ip = Ns*Is. Since the volt per turn RATIO must be the same on each side, the winding with higher turns has a higher voltage, or emf. Since the ampturns PRODUCT must be the same on both sides, the winding with the higher turns has the lower current, or mmf. It can't be any other way, as the CEL would be breached. Does this explanation make it clear? 


#5
Sep2707, 08:22 PM

P: 22

Basically, power in = power out. Now, power is potential times current, or volts x amps. So, if you have a 1000 VA transformer with a 100 V primary and a 500 V secondary, under load the primary will draw 1000/100 = 10 amps and the secondary load will be 1000/500 = 2 amps. Doing the reverse, 100 Volts x 10 amps = 1000 VA 500 Volts x 2 amps = 1000 VA Conservation of energy 


#6
Sep2707, 08:50 PM

P: 218

Based on your answers, I have the following notions regarding transformation. Are they correct?
When a changing emf is applied to the primary from the source (the rotating coil), a current is induced in the coil. This primary current produces the primary flux. This flux causes a self induced emf to develop in the primary, which opposes the applied emf. At all instants, the applied emf is equal to induced emf (considering primary to be a pure inductor with no ohmic resistance). Flux produced by the primary must be cancelled by flux produced by the secondary. Why? Because, the flux produced by primary induces an emf in the secondary and we know that current produced due to changing primary emf has the sole purpose of producing so much flux that the opposing emf so created wholly cancels the primary emf or the primary flux, which produced this secondary current in the first place(Lenz's law). So, the current which will be produced in the secondary must be such that it produces that much secondary flux which cancells the primary flux. Now, if max current in secondary is equal to max current in primary, then flux produced will be greater than the primary flux since secondary has more number of turns (flux=n * I). So current produced is less, only that much that the net flux is equal and opposite to primary flux. Let us now go in reverse direction. It is now clear that current produced will be less than in primary. What about the emf in secondary? Why will this emf be produced in the first place? Clearly, to produce a current that opposes this changing primary flux. The emf produced must be such that it produces the current discussed above. But as we know, the secondary will have its own induced emf, which this secondary emf has to cancel to let the above mentioned current run. Now, since the coil is longer than primary, L is larger for it (i.e. induced flux produced per ampere of current running through it is larger), which means induced emf produced will be larger (a rise of 10 turns per unit length in the inductor increases its induced flux per ampere by 100 h in an inductor of 1 m length, or so I think). Thus emf to be created for this current to run must be high. Thus, emf in sec is high and current is low. Sorry for the confusing explanations above. Mr V 


#7
Oct107, 11:57 AM

P: 218

I still have one confusion. The secondary flux is equal and opposite to primary flux (conservation of energy). Then why does current still flow in primary, when emf induced in secondary is opposing primary's emf.



#8
Oct207, 10:45 AM

P: 1,036

This B value links the secondary almost entirely. For a given B, f, A, and Ns, Vs is determined. Just as a fixed voltage determines the flux density value, a fixed flux density determines the induced voltage value. FL is bidirectional. Likewise with AL. When a load resistance is connected across the secondary, Is is induced in accordance with OL, Is=Vs/Rs. This current Is is accompanied by an mmf which counters the original emf, known as Lenz' law, or LL. This mmf would tend to reduce the flux density B, and the field intensity H. But remember that the primary flux (& secondary since the coupling is very good) is determined by the value of the CVS. A CVS forces a specific voltage value, and consequently a specific B & H value. When the primary voltage Vp is countered by the mmf of the secondary load, the CVS outputs a larger current whose mmf opposes the secondary mmf. In other words, just as Is and its mmf tend to reduce the flux and emf, Ip does just the opposite and restores them. Otherwise power would diminish greatly as soon as a load is connected. Does this make things clear? BR. Claude 


#9
Oct207, 10:50 AM

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P: 40,730

Good post by cabraham. I'd only add that for real AC voltage sources driving the primary, their output voltage will droop a bit, based on the extra current being drawn to drive the secondary load. Again, you can model the situation as having the secondary load resistance mirrored across the transformer, so that it is in parallel with the primary winding, ratioed by the square of the turns ratio.



#10
Jun309, 04:13 PM

P: 4,663

More on the original transformer question. When you apply a voltage to the primary connections, an EMF is developed across the primary coil. The current in the primary coil is 90 degrees out of phase with the primary voltage, because the ideal transformer with no load looks like an inductance (remember jwL). It is this primary coil current that generates the magnetic flux amplitude in the core. Faraday's law generates a voltage across the secondary coil. But the induced voltage is 90 degrees out of phase with the magnetic flux amplitude (remember dB/dt >jwB), and therefore the induced output voltage is in phase with the voltage across the primary coil. The magnetic flux amplitude in the core in phase with the input voltage is independent of the transformer load current because of Lenz's Law. The only magnetic flux in the core that is not cancelled by Lenz's law is the original excitation, due to the primary inductance, and is in phasequadrature with the input voltage. So this is why the magnetic flux excitation is independent (mostly) of the secondary load current. Think about it.



#11
Jun309, 11:38 PM

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P: 40,730

Just a quick note of explanation  a spam bot reopened this old thread earlier today, so I closed the thread while I dealt with the bot. But before I got it closed, Bob responded to the OP (not the bot) with useful information, so I want to let the thread stand reopened in case others want to add more information about transformer action.
Transformer action is not obvious, especially when you go beyond the simplistic model that we all learn first. I ran across a couple really good transformer modelling theory application notes from Midcom Transformer today at work, and I'll post links here tomorrow. 


#12
Jun409, 12:34 PM

P: 1,036

The magnetic flux, phi, in weber, divided by the core cross sectional area, Ac, in m^2, is the magnetic flux density B, in tesla. The amplitude of the core flux density B, is determined by the primary voltage, not primary current. Faraday's law, FL, is bidirectional. Just as a specific value of flux determines the voltage, so does a specific voltage determine a flux. To illustrate, consider a wound bobbin with the ferromagnetic core installed. A CVS of 120V, 60 Hz is connected across the primary. The flux density B is determined by Vp, the primary voltage amplitude, f, the frequency, Ac, the core area, and Np the primary no. of turns. Then, the value of magnetic field intensity H, is determined by the BH curve of the core material. Also, the magnetizing current is directly related to H via Ampere's law, AL. Now remove the core altogether. The value of B without the core is approximately the same as with the core, a little less since the magnetizing current increases greatly and I*R resistive drops increase resulting in less emf across the winding. But, without the core, H increases drastically as does Imag. Imag can increase by orders of magnitude. With time changing magnetic fields, current and voltage are BOTH needed since a time changing ENERGY is present. Time changing energy, dw/dt, is power. But power is I*V. So it is impossible to set up a time changing flux without BOTH Ip & Vp. Under static conditions a dc current has a dc magnetic flux w/o a voltage. But that is static. Under timevariation, it is impossible to have a flux unless Ip & Vp are BOTH NONZERO. This issue has been laid to rest in the 19th century. Is it Ip or Vp that creates the core magnetic flux? It is BOTH. No ferromagnetic material has been found with a permeability of zero, or infinite. Hence B & H cannot exist independently. Since B is in voltseconds per turnmeter^2, & H is in ampturns per meter, both Ip & Vp are needed for time varying flux. Does this help? Claude 


#13
Jun409, 12:40 PM

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P: 40,730

And here are the two application notes I mentioned from Midcom Transformer about modelling transformers:
http://www.midcominc.com/Tech/pdf/TN69.pdf http://www.midcominc.com/Tech/pdf/tn82.pdf . 


#14
Jun409, 03:39 PM

P: 4,663

When a transformer is powered under noload conditions, the primary inductance L_{1} determines the excitation current. Specifically, [1] I = V/jwL_{1}.Furthermore, we know that in the core (from Maxwell's equations) [2] curl H = J where H is the magnetic excitation (amp turns per meter), and J is current density. This may also be written [3] H = NI where NI is the coil ampturns per meter. In the iron, [4] B = u u_{0} H where u = relative permeability and u_{0} is the permeability of free space. so [5] B = u u_{0}N I. Thus the magnetic field B depends explicitly on the coil current I, and is in phase with I. In terms of transformer performance, if we plug [1] into [5] we get [6] B = u u_{0} N V/jwL_{1} So the magnetic field in the core is in phase quadrature with the applied voltage (power factor angle nearly 90 degrees), and the peak amplitude varies inversely with excitation frequency w. This is why transformers designed specifically for 60 Hz (where the peak fields are about 1.4 Tesla) will overheat at 50 Hz, where the peak fields would be (60/50) x 1.4 Tesla = 1.68 Tesla. Because the POWER required to excite a transformer under no load conditions depends on the power factor, the POWER is V*I = VI cos(90 degrees) = ~0. Summarizing, the only Maxwell's equation that determines the core excitation is curl H = J. 


#15
Jun409, 07:43 PM

P: 1,036

But I started with B = Vp/(4.44*f*Ac*Np), which is Faraday's Law, FL. Then, Vs = B*4.44*f*Ac*Ns, also FL. FL works both ways. But one can measure or compute the Imag value to get H, then times mu to get B, etc., which is what you did. It's equivalent. Remember that Lp = mu*Np^2*Ac/lc, where lc is the core's path length. The end result is the same. But my way is easier because the primary is generally driven from a CVS. If the source powering the primary was a constant current source, CCS, your method is more direct. For a current transformer, CT, your method is more direct. The core flux is established by both the current & voltage in the primary. Vp determines B, while Ip determines H. For a voltage xfmr, i.e. driven by a CVS, the B value is determined by Vp. Then Vs = (Ns/Np)*Vp. Imag doesn't matter, as well as Is. Transformers operate under 2 of Maxwell's laws, Ampere's law which you stated, curl H = J, and Faraday's law, curl E = dB/dt. Both are always in effect. Both methods give the same result, but FL gives a more direct answer for a VT, whereas AL is shorter computation for a CT. Both approaches give the correct result. Have I helped? Claude 


#16
Jun409, 10:57 PM

P: 4,663

Claude
The basic difference is that I use Maxwell's (physics) equations, and you use engineering equations. I took only two EE courses as an undergraduate, and one was an EE lab where we measured gain of amplidynes and transconductance curves of pentodes etc.. Bob S 


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