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When transformer secondary is short circuited, what would be the value of current flowing through short circuited side? Shouldn't it be infinity ideally? If yes, is it only due to large cross-section & highly conductive copper wires that wires don't get melted?

For e.g. suppose I have 1 kVA, single phase, 250/125 V transformer. Now if I short circuit 125V side and set full load primary current i.e. 1000/250 = 4 A in primary then what would be value of secondary current?

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  • \$\begingroup\$ Are you interested in what happens when you operate the transformer within its ratings, or what happens when you exceed its ratings? 4 amps, 250V input into a 1000VA rated transformer is no issue. But if you short circuit the secondary, then apply 250V to the input, you will get a lot more than 4 amps. \$\endgroup\$ – Li-aung Yip Mar 17 '15 at 8:07
  • \$\begingroup\$ Actually I was interested in both the cases. Your explanation is made it clear, thanks. \$\endgroup\$ – Deep Mar 17 '15 at 12:54
  • \$\begingroup\$ To add to your understanding - if you short circuit a transformer and apply full voltage to the input - the current will be limited not just by the resistance and inductance of the transformer, but also by the magnetic saturation of the iron core. \$\endgroup\$ – Li-aung Yip Mar 17 '15 at 12:56
  • \$\begingroup\$ Oh, I kinda forgot the core saturation thing. Thanks :) \$\endgroup\$ – Deep Mar 17 '15 at 13:03
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I have 1 kVA, single phase, 250/125 V transformer.

If you can adjust the primary voltage so that the input current is 4 amps RMS when the secondary is shorted, the secondary current will be 8 amps for a 2:1 step-down transformer.

NB - you will probably find that the easiest way to do this is use the output of a variac and raise the primary voltage carefully whilst noting the primary current. Typically the variac output will only need to be around 10V RMS to get full primary current with secondary shorted out.

The primary-secondary turns ratio determines both step-down voltage ratio and step-up current ratio.

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  • \$\begingroup\$ Thanks, but if transformer was ideal..I guess it would be just impossible to set primary current, am I right? And also by short-circuiting real transformer, we are actually, in one way, extra-overloading transformer, right? \$\endgroup\$ – Deep Mar 16 '15 at 16:26
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    \$\begingroup\$ You can set primary current by driving through an impedance such as a resistor ideal or non-ideal. If you short circuit the transformer, providing you are not drawing more than the full rated primary current then you won'nt be overloading it. This sort of test is a classic measurement on transformers and AC motors (locked rotor) in order to measure winding resistance and leakage inductances. \$\endgroup\$ – Andy aka Mar 16 '15 at 18:11

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