Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trying to model mutual inductance with losses

  1. Jan 17, 2017 #1
    So as a learning project I am trying to solve a transformer with losses but I am stuck.
    1 coil L1 with N1 turns is excited by a sinusoidal voltage V1 with a resistor R1 in series. Wound around a common core with a second coil L2 with N2 turns and a resistor R2 also in series.

    $$V_1+R_1I_1+L_1\frac{dI_1}{dt}-M\frac{dI_2}{dt}=0\\
    R_2I_2+L_2\frac{dI_2}{dt}-M\frac{dI_1}{dt}=0\\
    M\frac{dI_1}{dt} = N_2\frac{d\phi_{12}}{dt}\\
    M\frac{dI_2}{dt}=N_1\frac{d\phi_{21}}{dt}\\
    V_1+R_1I_1+N_1\frac{d\phi_{11}}{dt}-N_2\frac{d\phi_{21}}{dt}=0\\
    N_2\frac{d\phi_{22}}{dt}+R_2I_2-N_2\frac{d\phi_{12}}{dt}=0
    $$
    So I have the last 2 equations with 8 unknowns (phi self and mutual and I1 I2 and their derivatives)
    Now :
    $$\phi_{11} = \phi_{12}\\
    \phi_{21}=\phi_{22}$$
    I think this has to be the case, because all of the flux from the first coil that hits the second coil must also go through the first coil, since all the flux is confined to the core. So I will call them $$\phi_1\textrm{ and }\phi_2$$
    $$
    V_1+R_1I_1+N_1\frac{d\phi_{1}}{dt}-N_2\frac{d\phi_{2}}{dt}=0\\
    N_2\frac{d\phi_{2}}{dt}+R_2I_2-N_2\frac{d\phi_{1}}{dt}=0
    $$
    4 unknowns. Now I set up a magnetic circuit
    $$\textrm{Median length of core }= l\\
    \textrm{Permeability of core }= \mu\\
    \textrm{cross section of core }= A_x\\
    \textrm{Reluctance of core }(R_c)=\frac{l}{\mu A_x}\\$$
    $$N_1I_1=R_c\phi_1\\
    N_2I_2=R_c\phi_2\\
    \phi_1=\frac{N_1I_1}{R_c}\\
    \phi_2=\frac{N_2I_2}{R_c}$$
    I think I can just break the above 2 equations for the magnetic flux into 2 by superposition... Which gives 2 equations 4 unknowns:

    $$
    V_1+R_1I_1+N_1\frac{d}{dt}\frac{N_1I_1}{R_c}-N_2\frac{d}{dt}\frac{N_2I_2}{R_c}=0\\
    N_2\frac{d}{dt}\frac{N_2I_2}{R_c}+R_2I_2-N_2\frac{d}{dt}\frac{N_1I_1}{R_c}=0
    $$
    or
    $$
    V_1+R_1I_1+\frac{N_1^2}{R_c}\frac{dI_1}{dt}-\frac{N_2^2}{R_c}\frac{dI_2}{dt}=0\\
    \frac{N_2^2}{R_c}\frac{dI_2}{dt}+R_2I_2-\frac{N_2N_1}{R_c}\frac{dI_1}{dt}=0
    $$

    I think I am missing a step... because I think I1, I2 are dependent on each other somehow. What am I missing. Since this is not an ideal transformer, I don't think N1I1=N2I2.. so can someone tell me how I can relate them or what I did wrong please?
     
  2. jcsd
  3. Jan 21, 2017 #2
    At first the conventional voltage and current sense is different for supply circuit from supplied one.
    The supply circuit -generator- the emf and the terminal voltage are in the same sense. In the supplied circuit the supplied voltage is opposite sense the current and the counter emf.
    upload_2017-1-21_10-2-49.png
    So, the first equation has to be-in my opinion:
    v1=R1*i1+L1leakage*di1/dt+L1core*di1/dt-M*di2/dt
    In your case M12=M21=M~=SQRT(L1core*L2core)
    L2core*di2/dt-M*di1/dt+R2*i2+L2leakage*di2/dt=0
    L1leakage exists since part of the flux generated by i1 flows in air not in the core. The same at secondary coil.
    The magnetic flux in the core is considered constant. So if the secondary circuit is open- that
    means i2=0- then the flux in the core will be Io*Np/Rc [Rc=reluctance].When reclosing the secondary circuit the core flux stays the same Np*i1/Rc+Ns*i2/Rc=Np*Io/Rc
    There are a few ways to present the transformer magnetic circuits and the inductances in a transformer.
    In electric power engineering the Steinmetz Equivalent Circuit is employed.
    upload_2017-1-21_10-5-42.png

    Never the less you need here to know Rp,R’s,Xp,X’s and Xm and Rfe [the linkage reactance and the equivalent resistance of the core losses] and Vp-of course.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Trying to model mutual inductance with losses
Loading...