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I can't find formula for this. I have rated load current, acutal load current, rated output voltage and variability of output voltage (given in percentage).

How to find no-load output voltage and output voltage for given load current?

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  • \$\begingroup\$ How about sharing the data sheet? \$\endgroup\$ – Andy aka Feb 3 '17 at 22:00
  • \$\begingroup\$ It's not existing device, but academic homework. \$\endgroup\$ – Sławomir Kozok Feb 3 '17 at 22:08
  • \$\begingroup\$ Sn=680 kVA, U1n=20 kV, U2n=6,3 kV, fn=50 Hz, uZ%=6 %, ΔPCun=5 kW, ΔPFen=2,5 kW, U1=U1n, f=1,2fn \$\endgroup\$ – Sławomir Kozok Feb 3 '17 at 22:09
  • \$\begingroup\$ I only need one formula for output voltages. I've calculated another things needed in my homework. \$\endgroup\$ – Sławomir Kozok Feb 3 '17 at 22:10
  • \$\begingroup\$ The key value is uZ=6%. 1. You short the secondary 2. you increase the primary voltage until the the primary has the rated current. That's the short circuit voltage. uZ is given as the percentage of the rated primary voltage. Please look up the formulas for power transformer calculations in your textbook again. \$\endgroup\$ – Janka Feb 4 '17 at 0:12
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Rated Voltage at rated current for a source impedance of Z = 6% pu means that the source impedance is 6% of the rated load impedance so for rated current they allow a 6% series drop and short circuit current is 1/6% or 16.7 x rated current and no load voltage is 6% higher thus lower loads will be proportionally less

or given %Ipu, %Zpu , what is %Vpu?

%Vpu rise above rated voltage V load = Vrated + %Ipu * %Zpu

This is because %Zpu and %load regulation error are equivalent.

https://en.wikipedia.org/wiki/Per-unit_system

1 p.u. 100% per unit of rating for V,I,Z,P

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Calculate the Thevenin equivalent of the secondary output. From that, you can easily calculate the output with any particular load.

The open circuit output voltage is the Thevenin source voltage. With one other data point of voltage and current, you can compute the Thevenin source resistance.

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If it's academic homework, then go look up the equivalent circuit of a transformer, put numbers on all the reactances and resistances and back predict what the open circuit voltage is knowing the full load current and full load voltage. Strictly speaking you will need to know the complex nature of the impedance of the load that it is rated for. Here's a start for you: -

enter image description here

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  • \$\begingroup\$ Perhaps using the approximate circuit would be easier to calculate things and it's quicker as well. Hence, that's why we use the approximation circuit. \$\endgroup\$ – KingDuken Feb 4 '17 at 0:30
  • \$\begingroup\$ I'm not a big fan of compromise so may I suggest you offer up your viewpoint as an answer. \$\endgroup\$ – Andy aka Feb 4 '17 at 0:41
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The transformer equations for an ideal transformer with resistance are as follows:

$$ V_1 = R_1*I_1+L_1\frac{di}{dt}+M_s*I_1$$ $$ V_2 = R_2*I_2+L_2\frac{di}{dt}+M_s*I_2$$

Ms is the coupling coefficent.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ If you set the resistances to zero you can find the coupling for a real transformer. The equations are hard to find. \$\endgroup\$ – Voltage Spike Feb 3 '17 at 23:50
  • \$\begingroup\$ The units don't work out for \$Z_1\$ and \$Z_2\$. \$L\frac{\textrm{d}i}{\textrm{d}t}\$ has voltage units. You can't add that to units of resistance. \$\endgroup\$ – jonk Feb 4 '17 at 2:03
  • \$\begingroup\$ I had these in matrix form, translation error from my notes, is that better? \$\endgroup\$ – Voltage Spike Feb 4 '17 at 5:16
  • \$\begingroup\$ Yeah. Much better. Often, the first thing I do is dimensional analysis and stop looking further if that sniff test fails. \$\endgroup\$ – jonk Feb 4 '17 at 5:50

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