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I am confused about voltage polarity and current direction in transformers , for example the equivalent circuit of the ideal transformer in figure(a) the instantaneous voltage polarity of the secondary is positive at the dot terminal,if we connect a load across the seconday terminal, isn't supposed that the current goes out from the dot to the load? or does that mean that the voltage and current of the secondary side are 180 degree out of phase?

Equivalent circuits

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Note that in case (a) the ratio of the currents is negative while in (b) it is positive but the secondary current arrows are reversed. They both, effectively, say the same thing.

I've never seen it expressed as in (a) but I can see that it may make some sense to present an ideal transformer with current in from both sides as neither side is then assumed to be "input" or "output" but both can be inputs, etc.

(b) is the normal way of thinking in most electronics applications. You may find that (a) has its uses in electrical utility grid transformers where power can flow either direction to suit generation / demand requirements.

(c) and (d) should be fairly obvious inversions of (a) and (b) respectively.

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  • \$\begingroup\$ That's ok , but I still have the same inquiry , I will just copy the comment down : But if we derive the equation from the magnetic circuit shown here (with the aid of right hand rule) link the current ratio will be negative , so which physical configuration that make the current ratio positive as in case (b) \$\endgroup\$
    – iMohaned
    Mar 11 '16 at 23:20
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(b) is the conventional choice, with both \$i_1\$ and \$i_2\$ positive.

If you use (a), then one of the currents has to be negative, because the power flowing in one side has to equal the power flowing out the other.

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  • \$\begingroup\$ But if we derive the equation from the magnetic circuit shown here (with the aid of right hand rule) link the current ratio will be negative , so which physical configuration that make the current ratio positive as in case (b) \$\endgroup\$
    – iMohaned
    Mar 11 '16 at 23:16
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    \$\begingroup\$ That diagram corresponds to your figure (a), in which, as I said, one of the currents must be negative. All four diagrams above represent the same physical configuration. If you physically reverse the sense of one of the coils, you have to move its dot, too. The schematic diagram does not change. \$\endgroup\$
    – Dave Tweed
    Mar 11 '16 at 23:30
  • \$\begingroup\$ "All four diagrams above represent the same physical configuration." If I understood you correctly, then that's wrong; figures (a) and (b) are the same physical configuration, and (c) and (d) are the same, but (a)=(b) is not the same as (c)=(d). \$\endgroup\$ May 23 '20 at 0:59
  • \$\begingroup\$ @AlejandroNava: The fact that the dot has moved on the schematic does not mean that the physical construction of the transformer is any different. It is the dots that define how the transformer is constructed. The only thing that changes is the sign of the voltages and currents as denoted by the polarity symbols and arrows. \$\endgroup\$
    – Dave Tweed
    May 23 '20 at 3:46
  • \$\begingroup\$ Hi. All textbooks I've read so far say the other way around: you start with the transformer's construction, and to avoid showing the coils orientation in circuit diagrams since that's tedious, we use the dot convention instead. So, actually the coils orientation define where the dots will be placed. Anyways, that's not my point; cases (a)=(b) correspond to the case a) or b) of this image, and cases (c)=(d) to the case c) or d) of that image. \$\endgroup\$ May 23 '20 at 4:49
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The way you phrased the question is a bit confusing, so I'll just state how I think about the dot convention rule:

The voltage waveform on a dotted terminal is always in phase the voltage on another dotted terminal.

This is true only when comparing voltages to voltages - don't bring current into the mix. The phase relationship between voltage and current will depend on what else is attached to the transformer.

Also note that the instantaneous voltage is irrelevant - transformers only respond to time-varying voltages and currents, so it's best to think of any signal as a sine wave, not a voltage at a particular moment. If you need to focus on a particular moment, use dV/dt or di/dt.

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An ideal 1:1 transformer acts the same as a pair of wires, except that input is isolated from output. so if you cover the in picture b you can see current in the top an bottom wires matches and that voltage across input and output matches.

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You have to distinguish between a reference polarity or reference direction, which are arbitrary, and the numerical sign of the voltage or current, which depend on the chosen reference polarity/direction. It's easier to look at phasors (frequency domain) rather than the real instantaneous values (signals, time domain), but I'll use instantaneous quantities since your diagrams use them and you talked about instantaneous polarity.

[...] for example the equivalent circuit of the ideal transformer in figure(a) the instantaneous voltage polarity of the secondary is positive at the dot terminal,if we connect a load across the seconday terminal, isn't supposed that the current goes out from the dot to the load? or does that mean that the voltage and current of the secondary side are 180 degree out of phase?

With respect to phase angles, the transformer only relates the angles of the two currents, and the angles of the two voltages (for each pair, they're either in phase or 180° phase-shifted). For figure (a), just because the reference direction of \$i_2(t)\$ is entering the dotted terminal, it doesn't mean the actual current is entering the terminal for a given instant; you also need to consider the numerical sign at that instant (if it's positive, then the current is indeed entering, but if it's negative then the current is actually exiting positively). The equation \$ N_1/N_2 = v_1(t)/v_2(t) \$ says that the two voltages will be in phase; the equation \$ N_1/N_2 = - i_2(t)/i_1(t) \$ says that the two currents will have a phase shift of 180° for the given reference direction (both currents entering the dotted terminals). The equations don't say what will be the angle of the voltages or currents by themselves; in other words, if the phase of \$i_1(t)\$ is \$ \theta_{i1} \$, then the phase of \$i_2(t)\$ is \$ \theta_{i2} = \theta_{i1} \pm 180° \$; similarly, \$ \theta_{v2} = \theta_{v1} \$. This is what the transformer equation says regarding phase angles, however it doesn't indicate what's the numerical value of \$ \theta_{i1} \$ (and thus it doesn't indicate either the numerical value of \$ \theta_{i2} \$), therefore the transformer equation doesn't say whether the current in coil 2 is entering or exiting its dotted terminal. So, regarding your first question,

isn't supposed that the current goes out from the dot to the load?

the answer is no, in general the current can be exiting or entering the dotted terminal. Regarding your second question,

does that mean that the voltage and current of the secondary side are 180 degree out of phase?

the answer is no. The transformer doesn't relate the phase of \$i_2(t)\$ and \$v_2(t)\$; that's determined by the external circuit.

How do we know if \$i_2(t)\$ is actually exiting or entering the dotted terminal of coil 2? It depends on the rest of the circuit, the instant \$t\$ under consideration, and the angular reference. You need to give us a specific example so we can answer you correctly. But to guide you, once you know the current \$i_2(t)\$ (or its phasor \$\tilde I_2\$), then you can know whether the current is actually entering or exiting the dotted terminal, provided that you specify an instant (remember that in AC circuits, the actual polarity and direction of instantaneous voltages and currents are changing). For the sake of simplicity, let's assume we choose \$t=0\$, so \$i_2(t) = I_{m2} \cos ({\theta_{i2}}) \$. Looking at the graph of a cosine function, if \$\theta_{i2}\$ is between -90° and 90°, then \$i_2(t)\$ is positive and so it is indeed entering the dotted terminal; but if \$\theta_{i2}\$ is between -180° and -90°, or, 90°and 180°, then \$i_2(t)\$ is negative for the chosen reference direction and so it is actually exiting the dotted terminal.

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