transformer
If we apply a voltage at the primary terminals of a transformer then for this we can get a voltage at the secondary. But they have 180 degree phase difference. What is the reason behind this??
I want to know why the phase difference between primary voltage and secondary voltage is 180 degree.
What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -
Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -
\$V_P\cdot I_P = V_S\cdot I_S\$
The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.
This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.
The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.
Magnetizing inductance is in parallel with the perfect transformer model, not in series. Series inductance represents leakage inductance ie inductance that does not contribute to the magnetic field applied to the secondary winding..
Maybe you should try modelling both for a decent representation of reality. Then add on copper losses as series resistors and eddy current losses as a resistor in parallel with the primary. You should also consider self resonances due to interwinding capacitance.
Only then will you get a fair representation of reality.
Ignoring all the little effects of adding series and parallel resistance, inductance and capacitance, when you have 50:50 square wave going through a transformer you can forget about the DC content; the output will have an average value that is zero so, if the mark-space ratio were 10%, the output waveform would peak at 90% for 10% of the time and remain at -10% for 90% of the time.
If you need this to be tied to 0V so that at any reasonable MS ratio, the output voltage was clamped to round about 0V you'll need a series capacitor and a diode clamp to 0V. Now for a broad range of MS ratios, the peak voltage will be Vpk-Vdiode and the low voltage will be -Vdiode. Hope this helps.
Best Answer
Imagine you wound an inductor but instead of one thicker wire you used two thinner wires (this is commonplace by the way). An AC voltage is applied on both wires - lets call the AC voltage AC live and AC neutral and lets say neutral is connected to 0V for reference.
No problem so far?
Now, imagine you only applied AC live to one of those wires and left the other one open circuit (AC neutrals both at 0V still). What would be the induced voltage in the unconnected wire? Would it be: -
If the induced voltage on the unconnected wire were antiphase then joining it up to the AC live would create a seriously big problem and there would be sparks and breakers tripping. If the induced voltage on the unconnected wire were significantly different to the AC live and you joined them up then same problem as before - sparks and breaker trips.
This means if you wind a transformer with two wires in the same way there will be no phase shift but you can always swap the primary wires or secondary wires over and get 180 degrees.
Inherently there is no 180 degrees phase shift - you only get 180 degrees when you get into a muddle with the wires.