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We know that current transformer is used to step down the current of the primary side. So basically it is a STEP UP transformer. Also we say that current transformer basically works on close to short circuit mode.

So if \$ {V1\over V2} = {I2\over I1} = {N1\over N2} \$ and N2 >> N1 so V2 should be also very much high than V1. So if we are trying to short circuit this large voltage, how it is going to work as large current will flow in the secondary also. Don't you think it is going fundamentally wrong?

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2 Answers 2

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There's no contradiction. When you limit the secondary voltage by putting a load across it, the primary voltage becomes proportionally lower, too.

You can think of it as having the secondary load resistance "appearing" in parallel with the primary, but with a value that's reduced by a factor of \$(\frac{N_2}{N_1})^2\$. When you put a given current through this combined impedance, the voltage drop across the primary is much lower than it otherwise would be.

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  • \$\begingroup\$ just suppose the voltage across the primary is of considerable value... so the value of secondary voltage will be also large..now we are short-circuiting the scondary.. now how you going to justify what you said sir ?? \$\endgroup\$ May 23, 2015 at 18:43
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Simple worked example: -

The burden is (say) 10 ohms and the turns ratio might be (say) 500:1. This means the impedance seen by the primary is: -

\$\dfrac{10\space ohms}{100^2}\$ = 40 micro ohm

This transformed impedance is seen by current in the single-turn primary.

Of note is that this tiny resistance is in parallel with the transformer's magnetization inductance (let's say 100uH). At 50 Hz 100uH has an impedance of 31 milli ohms i.e. it represents a miniscule error.

What happens if the burden is removed? The magnetization inductance still offers quite a small impedance of a few milli ohms (reactive) so the universe will still be intact and, if 100 amps is flowing in the primary a voltage of about 1500 volts will be seen on the secondary.

This may cause problems with insulation breakdown on the secondary and might short out the secondary a bit. Net effect is a secondary short and this short, as seen by the primary, will make the primary look like a very small impedance indeed.

But remember this - the current in the primary is due to a voltage supply feeding a load - this remains largely (99.9%) of the case no-matter what you do with the secondary.

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  • \$\begingroup\$ just suppose the voltage across the primary is of considerable value... so the value of secondary voltage will be also large..now we are short-circuiting the secondary.. now how you going to justify what you said sir ?? \$\endgroup\$ May 24, 2015 at 5:34
  • \$\begingroup\$ If you can justify the circumstances of how the voltage on the primary can be considerable I will answer. \$\endgroup\$
    – Andy aka
    May 24, 2015 at 9:25
  • \$\begingroup\$ actually sir, my professor asked me this condition what will happen if the primary voltage is considerable. Even i somehow know that primary voltage is very less so is secondary voltage. If you can answer this please help me .. \$\endgroup\$ May 25, 2015 at 5:12
  • \$\begingroup\$ That is no justification. \$\endgroup\$
    – Andy aka
    May 25, 2015 at 9:31

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