Measuring short circuit current safely

Hi guys, I was just wondering how to measure short circuit current (or max current subject can supply) safely??
For example, I have a transformer with ratings 48V, 3A...

I wanna verify whether it really gives 3 Amps or not...How do i do that???:roll:

To do this test (measuring the transfomer output current) try on measuring the output voltage level of the transformer with a variable load connected. To simulate a variable load you may use some power resistor in parallel. Start connecting with a single power resistor to achieve let’s say 50% load current.
Calculate the value for
R1 = 2U/Imax R= 96/3= 32 ohms (33 ohms standard values), resistor rated for a maximum power of P= 1.2 (UI/2) = 86.4W (in practice use a 100W resistor)
Add supplementary resistors in small steps and keep monitoring the output voltage until voltage drop 10-15% from initial value on 50% load. That’s the point to consider a safe maximum output current for the transformer. Each supplementary resistor need to draw from the transformer up to 10% of the maximum current.
R2, R3, R4, R5, R6 = 10U/Imax = 160 ohms, (you may use standard value of 150 ohms), each resistor rated for a maximum power of Pn= 1.2 (UI/10) =17.28W (use 25W resistors)
Imax= U/Rx as Rx = R1||R2||R3||R4||R5||R6
That's a total of 5 steps to verify how the output voltage changes from 50% load current to 100% load current. :-D

If you cant afford a power resistor (they are quite expensive), the heating element of an old hair dryer will work just fine. It has a very small resistance which varies along the length of the wire, and it can obviously withstand a lot of power dissipation.

Short circuit current and maximum output current are obviously different numbers. Maximum output current usually refers to thermal ratings and can't be determined in a short term measurement. Short cicruit current as well as voltage drop can be calculated from Xs and Rs impedance parameters, you don't necessarily need to load the transformer with high currents to measure these parameters.

In power engineering, the usual method is to determine the impedance voltage (or short circuit voltage) as an equivalent parameter. The method shorts the secondary and adjusts a variable primary voltage until the nominal transformer current is achieved.

Thanks everyone for replying...
I do have another question : My multimeter says it can measure currents up to 10A....So why can't I simply set it to current mode and measure Isc.

...........I know, i get too silly sometimes...

The maximum short circuit current that can be obtained from the output of the transformer is limited by the impedance of the transformer and is determined by the multiplying the reciprocal of the impedance timed the full load current.
I _fault= I _nominal/% Impedance .
For example if your 48V transformer has a rating of 150VA and an impedance of 5% then the available fault current is
I _fault= 3 /.05= 60 A.
When a transformer sees large through faults they can get insulation damage based upon the heat created by the fault (depending on duration of the fault).