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I am trying to understand the working principle of current and voltage transformers.

I understand that V.T. and C.T. are basically just transformers, but I can't understand how secondary winding voltage and current are defined by the load on the secondary winding. Hence, for current transformer, the secondary winding is short circuted (very low impedence), and if roughly speaking, voltage on secondary is determined by the equation:

$$ U_{secondary} = U_{primary} * k; where\\k = \frac{Ns}{Np} $$

and current

$$ I_{secondary} = \frac{I_{primary}}k $$

Then the current transformer works as voltage step-up transformer and voltage should be some value, but instead, because of the low load impedence Z = 0. It follows from Ohm's Law

$$U = I * Z $$

that Usecondary is 0. Why do these equations contradict each other?

I want to know one equation that describes all parametrs of transformer where is possible to see how primary and secondary voltages and currents change depending on impedences in transformer.

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  • \$\begingroup\$ low impedance is not zero impedance. \$\endgroup\$ – Brian Drummond Jun 30 '14 at 9:43
  • \$\begingroup\$ Okay, then impedance can be as big if you connect Amper metr as load, I think it is pretty low. \$\endgroup\$ – Austris Jun 30 '14 at 10:03
  • \$\begingroup\$ There is no contradiction. Your first equation is still valid even if both voltages are zero. \$\endgroup\$ – Dave Tweed Jun 30 '14 at 10:11
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Why do these equations contradict each other?

First of all, you're using ideal transformer equations and that's fine as long as you apply them properly.

But you haven't applied them properly in this case.

Assuming an ideal transformer, if the secondary is loaded with a short-circuit, the voltage across the secondary is zero volts and thus, the primary voltage must be zero.

The fact is that the equations you provide must be satisfied simultaneously.

So, assuming the secondary is loaded with impedance \$Z\$, the transformer equation becomes

$$V_s = I_s \cdot Z = kV_p $$

but

$$I_s = \frac{I_p}{k} $$

thus

$$V_p = \frac{I_p}{k^2}\cdot Z$$

And there you have it, \$V_p\$ and \$I_p\$ are not independent of the secondary load \$Z\$.

In fact, you see that when \$Z = 0 \mathrm{\Omega}\$, the primary voltage must be zero for any finite primary current.

So, there's no contradiction.

I want to know one equation that describes all parametrs of transformer where is possible to see how primary and secondary voltages and currents change depending on impedences in transformer.

Then start with this model of a physical transformer

enter image description here

and solve for the primary and secondary voltages and currents.

If I recall correctly,

  • R1 models the primary winding resistance
  • X1 models the primary leakage inductance
  • Rm models core loss due to hysteresis
  • Xm models the finite permeability of the core
  • X2 models the secondary leakage inductance
  • R2 models the secondary winding resistance
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  • \$\begingroup\$ Thank you, I just was watching from wrong perspective, now its seems easy \$\endgroup\$ – Austris Jun 30 '14 at 20:37
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Well, ohms law is right... if you want to calculate voltage on transformer terminals - where you have short circuit - you will get 0V.

The formula \$U_{secondary} = U_{primary} * k\$ can be used to calculate voltage on transformer terminals when load resistance is much higher than transformer internal resistance. You cant use it for short circuit, because short circuit is definetly smaller resistance than a lot of thin copper wire.

If you want to investigate what is going on in a transformer under any conditions - you have to dig deeper and get familiar with transformer model. Here is real transformer simplified model - it applies to CT, VT and other typical transformers.

enter image description here

Image source: great Analog Devices library - maybe you should read that article

In that case - you can ignore capacitances. Consider Lprimary and Lsecondary as magnetically coupled coils. First "produces" magnetic flux, other "receives" that flux and produces electromotive force.

If you short circuit terminals 3 and 4 - whole electromotive force inducted on Lsecondary inductance will be present on R3 and R4 resistances and L3 and L4 inductances.

There will be also voltage drop on R1, R2, L1 and L2 because under higher load (or short circuit) there will be higher current on primary winding, and higher current causes higher voltage drop. Part of inductance responsible for magnetic flux creation will get smaller voltage. Smaller voltage will be used to produce magnetic flux, so electromotive force on magnetically coupled inductance Lsecondary will be smaller.

In case of not loaded transformer - you will "see" whole electromotive force between terminals 3 and 4, because at very low current you can ignore relatively small winding resistatances and parasitic reactances.

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Since a transformer moves power from one place to another, the single equation you're looking for, for a perfect transformer, is: Pout = Pin, where P is in watts.

Of course there's a little more to it than that, like for instance if you had a 240 volt hair dryer that dissipated 1000 watts, and you had a 24000 volt feed available, how would you transform that 24kV to 240V?

We know that the ratio of the voltages on the windings of a transformer will be equal to the ratio of the number of turns on the windings, so if we need to get 240V from 24000, the turns ratio will be 24000 : 240, or 100 : 1.

Similarly, since we know that the transformer transfers power from one place to another, and since the hair dryer will be using 1000 watts, the transformer must get 1000 watts from the 24kV feed and send it to the hair dryer.

Since the primary feed is at 24000 volts, and P = IE, then the current it must draw from the feed in order to get 1000 watts out of it is I = P/E = 1000W/24000V = 0.0417 amperes.

Notice now a curiously beautiful thing: Since the transformer is delivering 1000 watts to the hair dryer, and the hair dryer is operating at 240 volts, the current into it must be: I = P/E = 1000W/240V = 4.17 amperes, which is 100 times the current into the primary, which is equal to the inverse of the turns ratio of the transformer.

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