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Homework Statement
Hi. Why does one conduct an open-circuit and a short-circuit test on a transformer to determine its equivalent circuit parameters? I've read that a) the open-circuit test determines the parallel parameters - the R and X equating to the core loss, and b) the short-circuit test determines the series parameters - the R and X equating to the winding loss and leakage flux. 'Figure 1' on this website is the diagram I'm talking about;
http://claymore.engineer.gvsu.edu/~johnsodw/egr325mine/paper2/paper2.html
I've done this experiment in the lab, and have all the relevant raw data. I'm just looking for some understanding in why certain parameters can be ignored in the open-circuit case and must be considered in the shoty-circuit case, and vice versa.
Homework Equations
Open circuit:
Rm = V1^2/Pin
Xm = V1/Ir
Short circuit:
Req = Pin/I^2
Xeq = sqrt(((V2/I)^2)-Req^2)
The Attempt at a Solution
As I understand it, with an open circuit causing the secondary current to be zero, so should the primary current be zero. But there is a current and this is therefore the magnetising current which flows irrespective of load and is responsible for core loss and reactance. So from this you can work out Rm and Xm.
Under short circuit, the voltage across the secondary windings should be zero, but its not. So by applying a current, a voltage is dropped across the windings and this is used to measure its resistance. Also, some of the applied current to the primary winding is wasted in leakage flux (so its power will not be transfered) - and by measuring the secondary current you can find the difference between it and the current you would expect from an ideal transformer, and this difference is then the current causing the leakage flux?
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