Transformers

Step 1: determine positive direction for currents

1. select a positive direction for magnetic flux

2. the positive direction for currents is the direction in which the flux produced will be positive

Step 2

1. let the flux in the iron core be $\mathrm{\Phi }$

2. write down the equations of electromagnetic induction

$$\begin{array}{r}\begin{array}{rl}{\mathcal{E}}_1& =-\frac{d{\mathrm{\Phi }}_1}{dt}=-N_1\frac{d\mathrm{\Phi }}{dt}\\ {\mathcal{E}}_2& =-\frac{d{\mathrm{\Phi }}_2}{dt}=-N_2\frac{d\mathrm{\Phi }}{dt}\end{array}\end{array}$$

According to my last article on circuits,

$$\begin{array}{r}\begin{array}{rl}{U}_1& =-{\mathcal{E}}_1=N_1\frac{d\mathrm{\Phi }}{dt}\\ U_2& =-{\mathcal{E}}_2=N_2\frac{d\mathrm{\Phi }}{dt}\end{array}\end{array}$$

And this is where textbooks usually end. (except perhaps with another equation for COE)

$$U_1{I}_1=U_2{I}_2$$

Step 3: dive deeper

Let $L_i$ be the self-inductance of the $i$th coil, and $M$ be the mutual inductance.

Then

$$\begin{array}{r}\begin{array}{c}{\mathrm{\Phi }}_1=L_1{I}_1+M{I}_2\\ {\mathrm{\Phi }}_2=L_2{I}_2+M{I}_1\end{array}\end{array}$$

Thus

$$\begin{array}{r}\begin{array}{rl}{U}_1& =L_1\frac{d{I}_1}{dt}+M\frac{d{I}_2}{dt}\\ U_2& =L_2\frac{d{I}_2}{dt}+M\frac{d{I}_1}{dt}\end{array}\end{array}$$

Note

Mutual Inductance reciprocity theorem:

$$M=\frac{{\mathrm{\Psi }}_1}{I_2}=\frac{1}{I_2}\int {\mathbf{B}}_2\cdot d{\mathbf{a}}_1=\frac{1}{I_2}\int {\mathbf{A}}_2\cdot d{\mathbf{l}}_1=\frac{1}{I_2}\int (\frac{{\mu }_0}{4\pi }\int \frac{I_{2}d{\mathbf{l}}_{\mathbf{2}}}{r})\cdot d{\mathbf{l}}_1=\frac{{\mu }_0}{4\pi }\iint \frac{d{\mathbf{l}}_1\cdot d{\mathbf{l}}_2}{r}$$

$$M=\frac{{\mathrm{\Psi }}_1}{I_2}=\frac{{\mathrm{\Psi }}_2}{I_1}$$

Solving for ${\dot{I}}_1$ and ${\dot{I}}_2$, we get

$$\begin{array}{r}\begin{array}{rl}{\dot{I}}_1& =\frac{U_2{L}_1-U_{1}M}{L_1{L}_2-M^2}\\ {\dot{I}}_2& =\frac{U_1{L}_2-U_{2}M}{L_1{L}_2-M^2}\end{array}\end{array}$$

Switching to complex numbers[1] (by assuming a sinusoidal signal), we get

$$\begin{array}{r}\begin{array}{rl}j\omega {\tilde{I}}_{10}& =\frac{{\tilde{U}}_{20}{L}_1-{\tilde{U}}_{10}M}{L_1{L}_2-M^2}\\ j\omega {\tilde{I}}_{20}& =\frac{{\tilde{U}}_{10}{L}_2-{\tilde{U}}_{20}M}{L_1{L}_2-M^2}\end{array}\end{array}$$

Below is a visualization of the four voltages and currents in Geogebra. (unembedded version)

Step 4: Power

You may have noticed that there are two additional points on the graph. They are the complex powers.

Derivation:

For

$$\begin{array}{r}\begin{array}{rl}U& =U_0\cos (\omega t+{\phi }_U)\\ I& =I_0\cos (\omega t+{\phi }_I)\end{array}\end{array}$$

Power

$$\begin{array}{r}P=UI=U_0{I}_0\cos (\omega t+{\phi }_U)\cos (\omega t+{\phi }_I)\\ =U_0{I}_0\frac{1}{2}(\cos (2\omega t+{\phi }_U+{\phi }_I)+\cos ({\phi }_U-{\phi }_I))\end{array}$$

The first term is sinusoidal and averages to zero. It’s called reactive power. The second term does not average to zero and is the real power.

In complex form:

$$S_0=S=\frac{1}{2}\tilde{U}{\tilde{I}}^{\ast }=\frac{1}{2}{\tilde{U}}_0{\tilde{I}}_0^{\ast }=P_0+j{Q}_0$$

where the real part is the real power and the complex part is the reactive power.

---

[1]

We can always do so because $\mathrm{\Re }(x{)}^{\prime }=\mathrm{\Re }(x^{\prime })$