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Relating Transformer Parameter I2 to V1 (Non-ideal) Updated

  1. Jan 25, 2017 #1
    mod note: thread moved from technical forum so template is missing
    1. The problem statement, all variables and given/known data
    Relate secondary current to primary voltage in a single-phase transformer having resistive load for non-ideal situation.

    2. Relevant equations
    KVL, Voltage across resistor and inductor, ac voltage as a sine wave, reluctance, Faraday's law,other equations relevant to working principle of transformer.

    3. The attempt at a solution
    Good day everyone! I would like to verify whether my attempt for relating secondary current to primary voltage in a single-phase transformer having resistive load is correct or not. Are there other losses that I have not taken into account? If so, feel free to critique my work. This is actually a work of my colleagues and I for our research. So here's our go at it:


    Using KVL for the equivalent circuit in the primary side gives,


    [tex] v_1=v_{R_1}+v_{L_1} [/tex]


    Substituting their corresponding values gives

    [tex]v_{1max}{\mathrm{sin} \omega t\ }=i_1R_1+L_1\frac{di_1}{dt}[/tex]


    Solving DE for [itex] i_1 [/itex] gives

    [tex] i_1=\frac{v_{1max}}{R^2_1+{{\omega }^2L}^2_1}\left(R_1{\mathrm{sin} \omega t\ }-\omega L_1{\mathrm{cos} \omega t\ }\right)\ [/tex]


    Combining the sine wave and cosine wave and simplifying gives

    [tex] i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]


    Now since [itex] N_1i_1=\mathcal{R}\mathrm{\Phi }\ [/itex], solving for [itex] {\Phi } [/itex] gives

    [itex] \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ [/itex]. This is the flux that flows to the secondary circuit after applying reluctance of the core.



    Substituting [itex] i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/itex] to [itex] \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ [/itex] gives


    [tex] \mathrm{\Phi }=\frac{N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]

    To determine the flux linkage in the secondary winding, we multiply N2 to both sides, giving


    [tex] N_2\mathrm{\Phi }=\frac{N_2N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]


    Taking derivative wrt t gives the voltage across the secondary circuit. That is,


    [tex] v_2=N_2\frac{d\mathrm{\Phi }}{dt}=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ [/tex]


    Using KVL on the secondary circuit, we get


    [tex] v_2=v_{R_2}+v_{R_L}+v_{L_2}\ [/tex]


    Substituting their corresponding values give,


    [tex] \frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2R_2+i_2R_L+L_2\frac{di_2}{dt}\ [/tex]


    [tex] \frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2(R_2+R_L)+L_2\frac{di_2}{dt}\ [/tex]


    Solving DE for [itex] i_2 [/itex] gives,

    [itex] i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\left[{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2\right]}\left[\left(R_2+R_L\right){\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }+\omega L_2{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\right]\ [/itex]


    Combining the sine and cosine waves gives and simplifying gives,

    [tex] i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\sqrt{{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2}}{\mathrm{sin} \left\{\omega t+{\mathrm{arctan} \left[\frac{R_1\left(R_2+R_L\right)-{\omega }^2L_1L_2}{R_1\omega L_2+\left(R_2+R_L\right)\omega L_1}\right]\ }\right\}\ }\ [/tex]



    Critiquing this will be of great help for our group's thesis. Thank you so much. :)
     
    Last edited: Jan 25, 2017
  2. jcsd
  3. Jan 28, 2017 #2

    rude man

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    If L1 is the primary inductance then you're off to a bad start. Or you need to show a diagram of your non-ideal transformer, identifying these parameters.

    I would start with basic equations for your transformer:

    v1 = L1 di1/dt + M di2/dt
    v2 = M di1/dt + L2 di2/dt

    L1, L2 are primary & secondary inductances
    M = mutual inductance (have you considered this?)
    v1, v2 = pri & sec voltages
    i1, i2 = pri & secondary currents

    and work from there. These equations accommodate loading, leakage inductances, winding resistance, etc.

    If you want to include consideration of core fluxes see my Insight article at https://www.physicsforums.com/insights/misconceiving-mutual-inductance-coefficients/
     
  4. Jan 29, 2017 #3
    Thanks Sir rude man. Basing from your comment,

    v1 = L1 di1/dt + M di2/dt
    v2 = M di1/dt + L2 di2/dt


    This implies that this is a system of DE and I need to solve for the corresponding functions of i1 and i2 as functions of time, right? As for the v1 and v2, shall I express them explicitly in the form v = vmax sin(ωt + θ)? I think this would be impossible for i2 since I don't know its phase angle yet.

    M = mutual inductance (have you considered this?)
    Well Sir, you can see in the last equation the following term N1*N2/R which is equal to μ*N1*N2*A/l. Is this not equal to [itex]\sqrt{L_1L_2}[/itex]? I may be wrong. I don't know.

    "These equations accommodate loading, leakage inductances, winding resistance, etc."
    How are these inductances related to resistance? Is there an explicit formula expressing the relationship among these variables?

    P.S. I'm an Applied Mathematics student, not an Engineering one. But I'm studying this field for our research. Thanks in advance. :)
     
  5. Jan 29, 2017 #4

    rude man

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    Let's assume an input voltage V = V0sin(wt), then a series resistor R1 connected to the transformer primary v1, then a load resistor R2 across the secondary v2. Then the equations would be
    (V - v1)/R1 = i1
    v1 = L1 di1/dt + M di2/dt
    v2 = M di1/dt + L2 di2/dt
    v2 = -i2. (I hope this last answers your question about i2.)

    4 independent equations, 4 unknowns. Solve classically or, preferably, using phasors which I guess you haven't covered.
    . M = k√(L1 L2) and I don't see the "k" which is the coupling coefficient and affects mutual inductance. I believe you assumed unity coupling (all primary flux coupling into the secondary). This is close to true for a power transformer with its high-permeability core, but for (for example) an IF radio transformer it's far from true.

    I will try to find time to pay more attention to your rather extensive (and impressive!) post.
    Inductances L1 and L2 are the inductances of the primary & secondary windings respectively and have no direct relationship to the respective winding resistances.
    You can certainly use classical calculus to solve these equations, though phasors are preferred since they change the diff. eqs. to algebraic ones. But you do need at least basic circuit analysis to do anything with the equations I provided.

    Have you looked at my Insight paper? You need not be concerned with the main topic but I do relate electrical parameters to magnetic ones which may be of interest to you.
     
  6. Jan 29, 2017 #5
    For this set of equations,

    (V - v1)/R1 = i1
    v1 = L1 di1/dt + M di2/dt


    Are you trying to say that
    (V0sin(wt) - L1 di1/dt + M di2/dt)/R1 = i1, right? So V and v1 are actually different? If V is the input voltage, then what is v1?

    Also, I said this statement:

    I think this would be impossible for i2 since I don't know its phase angle yet.
    I'm actually referring to v2 instead of i2. I apologize for that.

    And for this equation:
    v2 = -i2
    Why is this so? I don't understand. I mean, they don't even have the same units. Are you referring to their respective magnitudes only?

    M = k√(L1 L2) and I don't see the "k" which is the coupling coefficient and affects mutual inductance. I believe you assumed unity coupling (all primary flux coupling into the secondary). This is close to true for a power transformer with its high-permeability core, but for (for example) an IF radio transformer it's far from true.
    Well, for this Sir, the result excluding k just surfaced from my derivations. Probably it is due to my assumption above that after reluctance is divided to the mmf, the resulting flux will link to the secondary side.

    Have you looked at my Insight paper? You need not be concerned with the main topic but I do relate electrical parameters to magnetic ones which may be of interest to you.
    I did, and you came up with the equation M = √(k1k2L1L2) which is pretty interesting. How do I know the values for k's?
     
  7. Jan 30, 2017 #6

    rude man

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    you're a bit sloppy with a sign here, but otherwise yes. V is the input voltage to resistor R1 in series with the transformer primary. v1 is the transformer primary voltage (the low end of both primary & secondary assumed grounded). Sorry I can't draw a diagram right now so go with what I described earlier.
    still don't see what your problem is . 4 equations, 4 unknowns. You are solving for all four of them and v2 is one of them.
    That was a big old blooper! I meant to write v2 = -R2 i2. R2 is a resistor across the secondary winding (the transformer load).
    yes I think that was the assumption which is fine for a power transformer and certain others having relatively high-permeability cores so that the flux is well confined to the intended magnetic path.
    That depends on how the transformer is manufactured. For most transformers k1 and k2 are the same or closly the same so there is just one coupling coefficient and then M = k√(L1 L2). For power transformers, even small ones, k ~ 1.0.
     
  8. Jan 31, 2017 #7
    Since you apparently want a steady state solution, there's no need to solve differential equations. Using the Laplace variable s, a mesh solution can be simply obtained:

    NonIdeal.png
     
  9. Jan 31, 2017 #8

    rude man

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    Knowledge or application of the Laplace transform is not necessary if the steady-state, sinusoidal solution is sought. Phasors suffice. But the OP is probably not familiar with phasors either, he/she stating that he/she is not an engineering but an applied math student. So classical calculus is probably his/her only way to go.
     
  10. Jan 31, 2017 #9
    Sirs, I already figured it out. But the equations really looked very hideous. I may be able to post it later.
     
  11. Jan 31, 2017 #10
    The Electrician, may I know what the initial equations looked like before they were set up in the matrix and before their Laplace transforms were taken? I also have a somehow considerable understanding of Laplace transforms and transfer functions btw, Sirs, but not that deep enough.
     
  12. Jan 31, 2017 #11
    There's no need to perform any Laplace transforms to perform steady state AC circuit analysis.

    See this page just beyond half way down, under the section heading "Generalized s-plane impedance" at this page: https://en.wikipedia.org/wiki/Electrical_impedance. The variable s can be taken to be simply jω for AC circuit analysis.

    Then read this page: https://en.wikipedia.org/wiki/Mesh_analysis for a short overview of the technique of phasor mesh analysis.

    The two mesh equations are:

    1.) (s*L1+R1)*I1 + (-s*M)*I2 = V1
    2.) (-s*M)*I1 + (s*L2+R2+RL)*I2 = 0

    The minus sign in the term "-s*M" is there because of the dotted ends of the transformer windings. If one of the dots were on the other end of the winding, the sign would be changed.

    In post #8, I showed the equations in matrix form, which allows the use of a matrix solver to obtain the solution. But any technique for solving a pair of simultaneous linear equations can be used.
     
  13. Jan 31, 2017 #12
  14. Jan 31, 2017 #13
    The Electrician, I now understand what you're trying to say and I tried so solve them manually. Luckily, I got the same answer for the magnitude and angle.
    If we translate this back in the t-domain, will i(t) = L^(-1){V1(s)}*(magnitude derived)*sin(wt + (angle derived)) or i(t) = V1max*magnitude derived* sin(wt + (angle derived)) or should the phase angle be negative? Thanks.
     
  15. Jan 31, 2017 #14
    The phase of the current with respect to the applied voltage is delayed, as you would expect in an inductor, but the current in the secondary of your transformer will experience a possible additional 180° phase shift depending on where the windings are dotted: https://en.wikipedia.org/wiki/Polarity_(mutual_inductance)
     
  16. Jan 31, 2017 #15
    How about if it is dotted the same way as you did in formulating the initial equations?

    And what will the equation in the time domain will look?
     
  17. Feb 1, 2017 #16
    Substitute some values for L1, L2, M, R1, R2, RL and ω into the expression I obtained for I2/V1:

    Tmp1.png

    and find the angle of the result. If the dot on one winding of the transformer is reversed, then substitute M = -M and re-evaluate the expression.

    This is your homework, not mine. What do you get for the time domain equation? I see a zero and a couple of poles in the expression, so I would expect a moderately complicated time domain expression.
     
  18. Feb 1, 2017 #17
    IMG_1486012782.791948.jpg

    I got this answer. Is this acceptable?
     
  19. Feb 2, 2017 #18
    Sir, also noticed that the answer is not in Amperes.
     
  20. Feb 2, 2017 #19
    Oops. I forgot the convolution theorem.
     
  21. Feb 2, 2017 #20
    This is not the steady state result. You will note that you have a decaying exponential factor. What you have is the result of suddenly applying a sine wave at t=0. Get rid of the exponential factor and the remaining part should be the steady state result.

    But you already have the steady state result in phasor format. Why have you bothered to invoke Laplace transforms?
     
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