Transformers#

transformer

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 \(\Phi\)

  2. write down the equations of electromagnetic induction

\[\begin{split} \newcommand{\emf}{\mathcal{E}} \begin{align*} \emf_1 & = -\frac{d\Phi_1}{dt} = -N_1\frac{d\Phi}{dt} \\ \emf_2 & = -\frac{d\Phi_2}{dt} = -N_2\frac{d\Phi}{dt} \\ \end{align*} \end{split}\]

According to my last article on circuits,

\[\begin{split} \newcommand{\emf}{\mathcal{E}} \begin{align*} U_1 & = -\emf_1 = N_1\frac{d\Phi}{dt} \\ U_2 & = -\emf_2 = N_2\frac{d\Phi}{dt} \\ \end{align*} \end{split}\]

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

\[ U_1I_1 = U_2I_2 \]

Step 3: dive deeper#

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

Then

\[\begin{split} \begin{align*} \Phi_1 = L_1I_1 + MI_2 \\ \Phi_2 = L_2I_2 + MI_1 \\ \end{align*} \end{split}\]

Thus

\[\begin{split} \begin{align*} U_1 & = L_1\frac{dI_1}{dt} + M\frac{dI_2}{dt} \\ U_2 & = L_2\frac{dI_2}{dt} + M\frac{dI_1}{dt} \\ \end{align*} \end{split}\]

Note

Mutual Inductance reciprocity theorem:

\[ \newcommand{\V}[1]{\mathbf{#1}} M = \frac{\Psi_1}{I_2} = \frac{1}{I_2}\int\V{B}_2\cdot d\V{a}_1 = \frac{1}{I_2}\int\V{A}_2\cdot d\V{l}_1 = \frac{1}{I_2}\int\left(\frac{\mu_0}{4\pi}\int\frac{I_2d\V{l_2}}{r}\right)\cdot d\V{l}_1 = \frac{\mu_0}{4\pi}\iint\frac{d\V{l}_1 \cdot d\V{l}_2}{r} \]
\[ M = \frac{\Psi_1}{I_2} = \frac{\Psi_2}{I_1} \]

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

\[\begin{split} \begin{align*} \dot{I}_1 &= \frac{U_2L_1 - U_1M}{L_1L_2 - M^2} \\ \dot{I}_2 &= \frac{U_1L_2 - U_2M}{L_1L_2 - M^2} \\ \end{align*} \end{split}\]

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

\[\begin{split} \begin{align*} j\omega\tilde{I}_{10} &= \frac{\tilde{U}_{20}L_1 - \tilde{U}_{10}M}{L_1L_2 - M^2} \\ j\omega\tilde{I}_{20} &= \frac{\tilde{U}_{10}L_2 - \tilde{U}_{20}M}{L_1L_2 - M^2} \\ \end{align*} \end{split}\]

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

\[\begin{split} \begin{align*} U &= U_0\cos(\omega t + \varphi_{U}) \\ I &= I_0\cos(\omega t + \varphi_{I}) \\ \end{align*} \end{split}\]

Power

\[\begin{split} P = UI = U_0I_0\cos(\omega t + \varphi_{U})\cos(\omega t + \varphi_{I}) \\ = U_0I_0\frac{1}{2}\left(\cos(2\omega t + \varphi_{U} + \varphi_{I}) + \cos(\varphi_{U} - \varphi_{I})\right) \end{split}\]

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:

\[ S_0 = S = \frac{1}{2}\tilde{U}\tilde{I}^* = \frac{1}{2}\tilde{U}_0\tilde{I}_0^* = P_0 + jQ_0 \]

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