Trying to model mutual inductance with losses

In summary, the magnetic flux in the core of a transformer is constant, but the flux in the secondary coil is not.
  • #1
fahraynk
186
6
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?
 
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  • #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.
 
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1. How do losses affect mutual inductance in a circuit?

When there are losses in a circuit, such as resistance, it causes a decrease in the overall mutual inductance. This is because some of the energy is lost as heat instead of being transferred between the two inductors.

2. Can mutual inductance be accurately modeled with losses?

While it is possible to model mutual inductance with losses, it is difficult to do so accurately. There are many factors that can affect the losses in a circuit, making it challenging to account for all of them in a model.

3. How do you calculate mutual inductance with losses?

To calculate mutual inductance with losses, you must first determine the total resistance in the circuit. Then, you can use this value along with the inductance values of the two inductors to calculate the total impedance and losses. This can be done using various mathematical equations and simulations.

4. Are there any techniques for minimizing losses in mutual inductance circuits?

Yes, there are a few techniques that can be used to minimize losses in mutual inductance circuits. One approach is to use low resistance materials for the conductors and to design the circuit in a way that minimizes eddy currents. Another technique is to use shielding or magnetic cores to reduce the effects of external magnetic fields.

5. How do you account for non-linearities in mutual inductance with losses?

Non-linearities in mutual inductance can be accounted for by using more complex mathematical models that take into account the non-linear behavior of materials and components in the circuit. This can be done through simulations and experiments, but it can be challenging to accurately model all the non-linearities in a circuit.

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