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A simple transformer consists of two windings, closely spaced and wrapped around a metallic core so that the flux through each winding is equal. One winding is driven with a sine-wave alternating current. Some unspecified load is connected to the secondary winding. The voltage across both windings is plotted on an oscilloscope.

According to Ampère's law, the magnetic flux through the primary coil is proportional to the current in the primary coil, so it will be a sine wave in-phase with the primary current.

According to Faraday's law of induction, the voltage induced across the secondary winding is proportional to the rate of change of the magnetic flux through it. But the derivative of a sine wave is another sine wave 90 degrees out of phase (or a cosine wave). Therefore, the secondary voltage should be 90 degrees out of phase with the primary voltage.

Why is this not the case?

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  • $\begingroup$ This is an electrical engineering question and the concept to look up is Mutual Inductance. Remind me tomorrow if no one else answers and I will try to give you a complete answer. $\endgroup$ – robert bristow-johnson Dec 13 '16 at 7:06
  • $\begingroup$ In the meantime, maybe read this. $\endgroup$ – robert bristow-johnson Dec 13 '16 at 7:17
  • $\begingroup$ @robertbristow-johnson It's the day after tomorrow. $\endgroup$ – user253751 Dec 15 '16 at 10:43
  • $\begingroup$ got it. working on a response now. $\endgroup$ – robert bristow-johnson Dec 15 '16 at 23:50
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    $\begingroup$ The point where you go wrong is that the flux through the core is not proportional to the current in the primary. The current in the secondary will also generate flux that passes through the core. $\endgroup$ – polwel Sep 5 at 12:00
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I will answer this as if this is an Electrical Engineering course or site. We electrical engineers use "$j$" for the imaginary unit: $j^2 = -1$. We use the small-case "$i$" to represent current in the time domain. We use small case for signals in the time domain (such as $v(t)$ and $i(t)$) and we use large case of the same letter (and subscript) to indicate the same signals in the frequency domain ($V(j\omega)$, $I(j\omega)$) or equivalently Laplace Transforms of the same signals ($V(s)$, $I(s)$).

And this is the convention for defining the voltages and currents in a two-port mutual inductance (otherwise known as a transformer). The dot convention $\bullet$ simply means that the two inductors, as seen looking at the wires marked with the dot are wound around the common core in the same sense. Let's say they both have the same "right-hand rule".

mutual inductance

Given these definitions and conventions, the time-domain volt-ampere equations for the mutual inductance is:

$$\begin{align} v_1(t) &= L_1 \frac{d}{dt}i_1(t) + M \frac{d}{dt}i_2(t) \\ v_2(t) &= M \frac{d}{dt}i_1(t) + L_2 \frac{d}{dt}i_2(t) \\ \end{align}$$

The mutual inductance, $M$, is the same value in both equations and is related to the two individual inductances, $L_1$ and $L_2$, as

$$ 0 \le M = k \sqrt{L_1 L_2} \le \sqrt{L_1 L_2} $$

$k$ is the "coefficient of coupling", which is always between 0 and 1.

$$ 0 \le k \le 1 $$

Also, given the dimensions of a simple cylindrical wound ideal inductor, the inductance is:

$$ L = \mu N^2 \pi r^2 \ell = \mu N^2 (\text{Vol}) $$

$\mu$ is the permeability of the core, $N$ is the number of turns of winding the wire (BTW, the wire must be at least thinly insulated, often they are "enameled") and $r$ and $\ell$ are the cross section radius and length of the cylinder. The cross-sectional area is $\pi r^2$ and the volume is $ \text{Vol} = \pi r^2 \ell $. The salient fact to remember is that the inductance is proportional to the square of the number of turns.

The Laplace Transforms of the two equations are:

$$\begin{align} V_1(s) &= s \, L_1 I_1(s) + s \, M I_2(s) \\ V_2(s) &= s \, M I_1(s) + s \, L_2 I_2(s) \\ \end{align}$$

or the frequency-domain representation is (substituting $s \leftarrow j\omega$):

$$\begin{align} V_1(j \omega) &= j \omega \, L_1 \, I_1(j \omega) + j \omega \, M \, I_2(j \omega) \\ V_2(j \omega) &= j \omega \, M \, I_1(j \omega) + j \omega \, L_2 \, I_2(j \omega) \\ \end{align}$$

I am not finding the perfect image on the web of an ideal transformer with a load and keeping with the current and voltage conventions I have above. So I am using this picture:

enter image description here

Now keeping in the Laplace domain, we can add two equations:

$$\begin{align} V_g(s) &= R_1 I_1(s) + V_1(s) \\ V_2(s) &= -I_2(s) R_2 \\ \end{align}$$

The reason why we show $-I_2$ is because current is defined as positive flowing into $L_2$ and that is opposite of (or negative of) the current flowing into $R_2$. And we don't really give a rat's ass about $V_g$.

So solving these three equations

$$\begin{align} V_1(s) &= s \, L_1 I_1(s) + s \, M I_2(s) \\ V_2(s) &= s \, M I_1(s) + s \, L_2 I_2(s) \\ V_2(s) &= -I_2(s) R_2 \\ \end{align}$$

for $V_1$ in terms of $V_2$ we get

$$\begin{align} I_2(s) &= \frac{-1}{R_2} V_2(s) \\ \\ I_1(s) &= \frac{V_2(s) - s L_2 I_2(s)}{sM} \\ &= \frac{V_2(s) + \frac{s L_2}{R_2} V_2(s)}{sM} \\ \\ V_1(s) &= s \, L_1 \frac{V_2(s) + \frac{s L_2}{R_2} V_2(s)}{sM} + s \, M \frac{-1}{R_2} V_2(s) \\ &= \left(\frac{1}{k}\sqrt{\frac{L_1}{L_2}} + s \left(\frac{1}{k} - k \right)\frac{\sqrt{L_1 L_2}}{R_2} \right) V_2(s) \end{align}$$

Now, in the ideal case, when the coefficient of coupling, $k$, approaches 1, then the second term on the right goes to 0 and we have

$$ V_1(s) = \sqrt{\frac{L_1}{L_2}} V_2(s) $$

and $\sqrt{\frac{L_1}{L_2}}$ is real and positive and no $s$ factor in it. If the two coils are interwound over the same core and share the same dimensions (same $r$ and same $\ell$) then the square root of the ratio of inductances is the same as the turns ratio of the primary-to-secondary windings which is $\frac{N_1}{N_2}$. So

$$ V_1(s) = \frac{N_1}{N_2} V_2(s) $$

After inverse Laplace transforming, we have

$$ v_1(t) = \frac{N_1}{N_2} v_2(t) $$

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