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My book has the following passage which I am having trouble understand. I tried to search more about the topic on the internet and in other books but found none.

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Let the flux through each turn be $\phi$. The emf induced in the primary is $-N_1\frac{d\phi}{dt}$ and that induced in the secondary is $-N_2\frac{d\phi}{dt}=E_2$. If we neglect the resistance in the primary circuit, KVL gives us,

$$E_1-N_1\frac{d\phi}{dt}=0 \tag{i}\label{eq1}$$

Also, $$E_2=-N_2\frac{d\phi}{dt} \tag{ii}\label{eq2}$$

Thus, $$E_2=-\frac {N_2}{N_1}E_1$$

Let us first consider the case when the terminals of the secondary are not connected to any external circuit. The secondary circuit is incomplete and the current through it is zero. Suppose, the current in the primary is $i_s$ in this case (the subscript $s$ stands for the source and not for the secondary). As we have neglected the resistance in the primary circuit, it is a purely inductive circuit. The current has a phase difference of $90^\circ$ with the applied emf $E_1$, and hence the power delivered by the AC source is zero. The power in the secondary circuit is anyway zero as there is no current in this circuit.

Now suppose, the terminals of the secondary are joined to a resistance $R$. There will he an alternating current $i_2$ through R. There will be additional emf's induced in the primary as well as in the secondary due to $i_2$. But the net induced emf in the primary should remain equal and opposite to the source-emf $E_1$, by (i). So, there will be an additional current $i_1$, in the primary circuit which will cancel the emf induced due to $i_2$. Thus the primary will be $i_s+i_1$ and in the secondary $i_2$. The emf in the secondary will remain $E_2$ as given by (ii).

As the induced emfs due to $i_1$ and $i_2$ always cancel each other, the two alternating currents should be $180^\circ$ out of phase. Also $i_2$ is in phase with $E_2$ (purely resistive circuit), and $E_1$ is $180^\circ$ out of phase with $E_2$. This shows that $i_1$ is in phase with $E_1$ (figure below)

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My doubts are:

  1. What causes these additional emfs? Is it self induction?
  2. How does $i_1$ cancel out the induced emf due to $i_2$?
  3. Why is the necessary for $i_1$ to cancel out the induced emf due to $i_2$?
  4. Why is $i_1$ in phase with $E_1$? Here is my reasoning:

    $E_1$ and $E_2$ are $180^\circ$ out of phase. We take $\phi'$ as the induced flux due to $i_2$.

    Now, $$E_2-N_2\frac{d\phi'}{dt}=i_2R \\ \Rightarrow E_2-i_2R=N_2\frac{d\phi'}{dt}$$ $E_2$ and $i_2$ are in phase with each other. Hence $\frac{d\phi'}{dt}$ is in phase with $E_2$.

    Taking $\mathscr{E_3}$ as the induced emf in primary circuit due to $i_2$, we get $$\mathscr{E_3}=-N_1\frac{d\phi'}{dt}$$ Hence $\mathscr{E_3}$ is $180^\circ$ out of phase with $E_2$ and in phase with $E_1$.

    To cancel out $\mathscr{E_3}$, and emf opposite to its phase must be produced by $i_1$. Hence $i_1$ should be $180^\circ$ out of phase with $E_1$, which is not what the book says.

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1 Answer 1

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What causes these additional emfs? Is it self induction?

The current in the primary creates flux in the core. Changes in the flux create emf in both coils.

The current in the secondary creates "additional" flux in the core (but the direction of this flux will be opposite to the direction of the flux produced by the primary current, leading to less total flux, not more). And changes in this flux also create emf in both coils.

How does $i_1$ cancel out the induced emf due to $i_2$?

Because the directions of the fluxes produced by the two currents are opposite, the emfs they generate are also opposite.

Why is the necessary for $i_1$ to cancel out the induced emf due to $i_2$?

One way to look at it: If they didn't cancel out, then you'd have both energy being stored in the magnetic field in the core, and being dissipated as heat in the resistor. Energy wouldn't be conserved.

Another way to look at it: If they didn't cancel out, $E_1$ (the emf in the primary coil) would not be equal to $\mathscr{E}_1$ (the emf or voltage of the generator) and Kirchhoff's voltage law would be violated.

Why is $i_1$ in phase with $E_2$? The emf induced by $i_2$ in primary circuit should be $180^\circ$ out of phase with $E_2$ or in phase with with $E_1$. Hence should not $i_1$ be $180^\circ$ out of phase with $E_1$ to be able to cancel the induced emf?

Your quoted text is not distinguishing between $E_1$ (the potential difference across the primary) and $\mathscr{E}_1$ (the emf or potential diference across the generator). At one point you even claim the source text says, "$E_1$ is $180^\circ$ out of phase with $E_1$".

Whether $i_1$ is in phase or $180^\circ$ out of phase with $E_1$ depends what reference polarity you chose for $E_1$, something you haven't shared in your question.

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  • $\begingroup$ Thank you. That was helpful. Could you recommend me some resources where I could learn more about this topic? $\endgroup$ Commented Feb 20, 2022 at 1:59
  • $\begingroup$ I edited the fourth question a bit to make it clearer and fixed the transcript error in the quoted text. Could you take a look? $\endgroup$ Commented Feb 20, 2022 at 3:05

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