# Transformer Self Induction

1. Jan 4, 2013

### Ezio3.1415

I am having trouble thinking about self induction in the primary coil... My text says in standard situation the self induction voltage is same as the primary voltage... My question is how does current flow in the primary coil then? Please tell me where I am getting this wrong?

2. Jan 4, 2013

### Staff: Mentor

Well, I can't claim to be an expert, but I'll try a go at this. Forgive me if this doesn't make sense.

Inductors oppose changes in current. When current first starts to flow the magnetic field that it develops across the inductor cuts the windings and induces an opposite voltage. However this is only temporary. The opposition drops off as current increases until the current flow is steady, at which time there is no self inductance anymore until the current begins to decrease.

If we cut this into an infinite amount of infinitely small time steps we would see current flow start, which creates the magnetic field, which then cuts across the windings and opposes this increase in current. This opposition stops the increase in current, which means the magnetic field stops building, which stops the counter EMF and then allows the current to increase once more. Rinse and repeat.

Of course there isn't really any time steps, this all happens rather smoothly at the same time.

3. Jan 4, 2013

### BruceW

It sounds fine. I think the idea is that in the 'primary circuit' or whatever it is called (sorry I don't know much electronics terminology), there is almost zero resistance. Therefore, the input voltage must match the voltage across the primary coil due to inductance. Remember, the total voltage around a closed loop must equal zero. And since we assume almost zero resistance, there is no IR term. But just because there is no IR term, doesn't mean I=0

4. Jan 4, 2013

### jim hardy

Good observation on zero resistance...

Do they still teach this while still assuming sine wave voltage? Sine waves are a special case, mathematically.

Does it help you to think of voltage as a sine function, and current as ∫sin ?
Draw both on a scrap of graph paper, time on horizontal like an oscilloscope..
Observe that over course of a cycle they aren't always same polarity .
From that you can visualize voltage sometimes aiding and sometimes opposing current flow....
so their product(power) alternates positive and negative. That's saying energy cycles back and forth between source and inductor...
That's why no heat is dissipated in the inductor it's just swapped to somewhere else..
It helped me to draw a graph of volts, amps and power over a complete sinewave cycle, observing the above...

It would sound very unscientific to say
"...cosine shaped current sneaks in while sine shaped voltage aids it, and is yanked back out by nape of neck while opposed...",,,
but if it helps you past this stumbling block you dont ever have to admit you stooped so low..

Key point is this - induction doesn't oppose current, it opposes Change of current di/dt,,
AND Sinewaves are that special case where f(t), f'(t) and ∫f(t) all have exact same shape.
So it's not intuitive what is going on.

If this just clouds the matter, disregard. But it helped me.

old jim

Last edited: Jan 4, 2013
5. Jan 5, 2013

### vanhees71

I don't understand the question really, because I don't know what you mean by "voltage". There is no voltage here, because for a transformer to have a non-trivial function you have AC and thus a time varying magnetic field and thus, according to one of the fundamental laws of electromagnetism (Faraday's Law) a non-vanishing curl of the electric field
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
To answer your question concerning inductance in quasistationary circuit theory we integrate this equation over an arbitrary area $A$ with boundary $\partial A$ (assuming that this area and its boundary are at rest). According to Stoke's Law you have
$$\int_{\partial A} \mathrm{d} \vec{x} \cdot \vec{E}=-\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d} \vec{A} \cdot \vec{B}=-\frac{\mathrm{d} \Phi}{\mathrm{d} t}.$$
Now we assume that we have $n$ circuits with currents $i_k$ running around. To get the magnetic field due to the $k^{\text{th}}$ circuit we note that $\vec{\nabla} \cdot \vec{B}=0$ and thus $\vec{B}=\vec{\nabla} \times \vec{A}$, where we can use the Coulomb gauge constraint $\vec{\nabla}\cdot \vec{A}=0$. Then, neglecting the displacement current (which is justified if the typical frequency of the currents $\omega$ is such that $\lambda=\frac{2 \pi c}{\omega} \gg d$, where $d$ is the typical geometrical extension of the circuits) we have
$$\Delta \vec{A}=-\mu \vec{J},$$
$$\vec{A}(\vec{x})=\frac{\mu}{4 \pi} \int \mathrm{d}^3 \vec{x}' \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
The magnetic flux through the $k^{\text{th}}$ circuit thus is
$$\Phi_k=\int_{A_k} \mathrm{d} \vec{A} \cdot \vec{B}=\int_{\partial A_k} \mathrm{d} \vec{x}_k \cdot \vec{A}(\vec{x})=\int_{\partial A_k} \mathrm{d} \vec{x}_k \frac{\mu}{4 \pi} \sum_{j=1}^{n} \int_{V_j} \mathrm{d}^3 \vec{x}_j \frac{\vec{J}_j(\vec{x}_j)}{|\vec{x}_k-\vec{x}_j|}.$$
Now we assume that the wires are thin, i.e., that the current density $|\vec{J}_j|=i_j/C_j$, where $C_j$ is the cross-sectional area of the $j^{\text{th}}$ wire. From this we see that there must be coefficients $L_{kj}$ such that
$$\Phi_k=\sum_{j=1}^{n} L_{kj} i_j.$$
The $L_{kj}$ are the induction coefficients. It's easy to give the coefficients for $j \neq {k}$, because then we can write
$$\mathrm{d}^3 \vec{x}_j \vec{J}_j=i_j \mathrm{d} \vec{x}_j$$
and we find
$$\Phi_{k}^{(j)}=L_{kj} i_j=i_j \int_{\partial A_k} \mathrm{d} \vec{x}_k \cdot \int_{\partial A_j} \mathrm{d} \vec{x}_j \frac{\mu}{4 \pi |\vec{x}_j-\vec{x}_k|}.$$
The double integral is thus just $L_{kj}$.

We cannot use this formula for $L_{kk}$, because then we get into trouble with the singularity if the denominator of the expression under double-line-integral. Then we rather have to use $\mathrm{d} \vec{x}_k=\frac{1}{i_k} \mathrm{d}^3 \vec{x}_k \vec{J}_k(\vec{x}_k)$ in the line integral for $\Phi_k$. This gives
$$L_{kk}=\frac{\mu}{4 \pi i_k^2} \int \mathrm{d}^3 \vec{x}_k \int \mathrm{d}^3 \vec{x}_k' \frac{\vec{J}_k(\vec{x}_k) \cdot \vec{J}_k(\vec{x}_k')}{|\vec{x}_k-\vec{x}_k'|}.$$
In this integral the singularities are cancelled by the volume elements $\mathrm{d}^3 \vec{x}_k$ and $\mathrm{d}^3 \vec{x}_k'$.

In any case we can think the circuit to be described simply by the induction coefficients $L_{kj}$ together with their resistance (and capacities if there are capacitors involved, but let's keep the story simple by excluding this possibility for now). Then we go back to Faraday's Law in its integrated form. For the left-hand side we need
$$\vec{E}_k = \vec{J}_k/\sigma_k+\vec{E}_{k,\text{source}}$$
to get
$$\int_{\partial A_k} \mathrm{d} \vec{x}_k \cdot \vec{E}_k=R_k i_k-U_{k}.$$
Here $U_k$ is indeed a voltage (or better said potential) of the electric field from external sources. We can consider it as a given function of time.

Then Faraday's Law gives the following coupled set of equations
$$U_k=R_k i_k+\sum_{j=1}^{n} L_{kj} i_j.$$
Of course, there are more complicated cases, where the circuits are not only coupled by their mutual induction via the magnetic fields but in addition also connected via resistors, and/or capacitors. Then one also needs Kirchhoff's first law, which immediately follows from
$$\vec{\nabla} \cdot \vec{J}=0$$, which is approximately valid in the here considered quasistationary limit. It leads to the rule that at each branching in the circuit the sum of all currents must be 0.

6. Jan 7, 2013

### Ezio3.1415

"induction doesn't oppose current, it opposes Change of current di/dt,," This is the point...

And Drakkith's explanation I think simply explains why the same voltage isn't stopping the current...

And another point I didn't understand...
"The primary coil has zero resistance... Therefore, the input voltage must match the voltage across the primary coil due to inductance..."
I thought self induction depends on only the change of flux... What does it have to do with resistance of the primary coil?

7. Jan 7, 2013

### BruceW

If we assume a given input voltage (in other words, primary voltage), and assume the primary coil has some resistance, then the self induction will be smaller than the primary voltage. In fact, the 'time-average magnitudes' (for lack of a better word), of the primary voltage and the self-induction are related by this equation:
$$V_{selfinduction} \sqrt{1+ \frac{R^2}{L^2 \omega^2}} = V_{primary}$$
Where omega is the (angular) frequency of the primary voltage. You can see from the equation that the self-induction will be less than the primary voltage, unless the resistance in the primary coil is zero, in which case, they are equal. (Of course, the equation will get a bit more complicated if a large enough current is induced in the secondary coil).

So, when your text says "in standard situation the self induction voltage is same as the primary voltage" They mean that the standard situation is that the primary coil has effectively zero resistance.

What I was trying to explain in my first post is that since the total change in voltage around a loop is zero, we can think of the total voltage changes around the primary 'circuit'. Now, if there are no resistance in the primary circuit, then it is super-simple, because the only voltage changes as you go around the loop are due to the primary voltage and the self-inductance. Therefore, the primary voltage and self-inductance must cancel each other out.

So this gives us the reason why the self-inductance and primary voltage are the same (assuming there are zero resistance, and zero capacitance, in other words, that the primary coil is an ideal inductor).

Last edited: Jan 7, 2013
8. Jan 8, 2013

### Ezio3.1415

Now I understand your point... Thank you...