The current transformer secondary is short circuited during the operation. What is the reason, and explain with details.

Unless a burden (i.e. meters, relays, etc.) is connected to the CT, current transformers should always be shorted across the secondary terminals. The reason is very high voltages will be induced at the terminals. Think of the CT as a transformer, with a 1 turn primary and many turns on the secondary. When current is flowing through the primary, the resulting voltage induced in the secondary can be quite high, on the order of kilovolts. When a CT fails under open circuit conditions, the cause of failure is insulation breakdown, either at the shorting terminal strip, or at the feedthrough (in the case of oil filled apparatus), because the distances between terminals are not sufficient for the voltages present.

The next question is how does it work so by shorting the secondary terminal of CT, there's no over-voltage happen?

Because you typically develop little voltage across a short or the reflected impedance of a short.

Regards

cww

Regards

cww

CT is connect in series with the load, when CT is shorted or connected to certain load, the impedance become negligible compared to circuit its connected, so voltage drop across CT will be negligible.

When CT gets open circuited, the impedance of CT become infinity, hence the voltage drop across CT primary will try to rise up to to the rated supply voltage, but in this process insulation fails.

When CT gets open circuited, the impedance of CT become infinity, hence the voltage drop across CT primary will try to rise up to to the rated supply voltage, but in this process insulation fails.

CT secondary will develop a very high voltage when secondary is open circuited with the primary energised. But if the CT secondary is kept open initially and the primary is energised, will the CT get damaged?

It is all flux theory of transformer taught us in school, i.e. if open secondary, no opposition to primary flux and increases tremendously.

> CT secondary will develop a very high voltage when secondary is open circuited

> with the primary energised. But if the CT secondary is kept open initially and

> the primary is energised, will the CT get damaged?

of course, definitely the coils of CT will get damaged.

CT primary energised also secondary energised, but open circuited what will happen. it will burn.

> with the primary energised. But if the CT secondary is kept open initially and

> the primary is energised, will the CT get damaged?

of course, definitely the coils of CT will get damaged.

CT primary energised also secondary energised, but open circuited what will happen. it will burn.

> CT secondary will develop a very high voltage when secondary is open circuited with the primary energised. But if the

> CT secondary is kept open initially and the primary is energised, will the CT get damaged?

Depending on how long the secondary is kept open, not necessarily. This is exactly how Spark Plugs work in a car, and have worked for... well almost forever. The Secondary coil of the ignition coil, or the distributor, opens, causing a high voltage build up in the Primary, When the Secondary closes again,the extremely high voltage transfers to the secondary, and finally across the spark plug gap to ground, causing the spark in the spark plug, and that happens to the spark plugs/coil thousands of times a minute in some cases.

> CT secondary is kept open initially and the primary is energised, will the CT get damaged?

Depending on how long the secondary is kept open, not necessarily. This is exactly how Spark Plugs work in a car, and have worked for... well almost forever. The Secondary coil of the ignition coil, or the distributor, opens, causing a high voltage build up in the Primary, When the Secondary closes again,the extremely high voltage transfers to the secondary, and finally across the spark plug gap to ground, causing the spark in the spark plug, and that happens to the spark plugs/coil thousands of times a minute in some cases.

why voltage not high when we short the C.T secondary side?

Pls explain with principle.

Pls explain with principle.

CT secondary current is based on primary current (based on load). As per CT design, when CT shorted condition voltage is limited, if CT is open current is zero, voltage will be very high, during this time CT will burnt.

Basically a ct is a step up transformer, normally in the primary of a ct high currents are flowing. If the secondary is opened then as it's a step up transformer.

A high voltage is induced in the secondary, the induced secondary voltage is very high, it may damage the instrument itself or it may damage the observer or the insulation may break down so it must be shorted with an ammeter of suitable range or must be shorted.

A high voltage is induced in the secondary, the induced secondary voltage is very high, it may damage the instrument itself or it may damage the observer or the insulation may break down so it must be shorted with an ammeter of suitable range or must be shorted.

As we know that, CT works as step up transformer because the voltage induced across the secondary terminal is so high in terms of Mv. So if the secondary will open, its insulation can not bear the pressure of voltage it will fail, due to that fire will occurs. So during the installation time, secondary of CTis short circuited. Due to short circuit the voltage across the secondary will become zero and current become maximum, and we know that the current is very low in this case.

actually the voltage ratio is just opposite to the current ratio means: if there is 1000 amp in the primary side then it would be 1000 on the secondary also 1000/1,rather being n a step down. X-mer so the ends are short circuited !!!

> The current transformer secondary is short circuited during the operation.

> What is the reason, and explain with details.

(CT) is a type of instrument transformer specially designed for protection and measuring purpose, it will operate by the current whenever it exceeds the set value. The C.T converts the 1000A/5A. In the secondary more tappings we will take depend upon the measurement. then Electrons will flow in the closed path only. So the secondary will short circuited.

> What is the reason, and explain with details.

(CT) is a type of instrument transformer specially designed for protection and measuring purpose, it will operate by the current whenever it exceeds the set value. The C.T converts the 1000A/5A. In the secondary more tappings we will take depend upon the measurement. then Electrons will flow in the closed path only. So the secondary will short circuited.

> The current transformer secondary is short circuited during the operation.

> What is the reason, and explain with details.

Most replies answered by pointing out that destructive voltages will be generated if the secondary is open-circuited. While this is true, it does not exactly answer the question.

A current xformer will still operate safely over a wide range of secondary load resistances, so why is a very low secondary load resistance (= approximately a short circuit) the best choice?

The reason is that the fidelity of the current transformer is highest when the secondary is short circuited.

The purpose of a current xformer is to provide a scaled-down version of the primary current with the highest possible fidelity, that is, the secondary current should be a faithful replica of the primary current. In many applications it is not just the magnitude of the measured current that is important, but also faithful reproduction of the phase and high-order harmonics.

Thus, the question becomes, why is the fidelity highest when the secondary is shorted? At first this seems counter-intuitive, for the fidelity of the more familiar voltage transformer is most definitely NOT optimised by shorting the secondary! Indeed, as a voltage transformer is loaded more heavily by reducing the secondary load resistance, the secondary voltage 'sags', phase error is increased, and frequency response decreased, all leading to a loss of fidelity of the secondary voltage waveform. This occurs mainly due to winding resistance and leakage inductance, which would both be zero in an 'ideal' xformer. For a voltage transformer, the fidelity of the secondary voltage is highest with a high secondary load resistance, which draws very little current.

However, with current xformers, just about everything turns out to be reversed compared to voltage xformers, which can be confusing. As with all xformers, the secondary load resistance is reflected back to the primary side, scaled by the square of the turns ratio. Therefore, if we short the secondary, then we also short the primary. Note that the primary current is fixed and constant, being set by the external circuit. Therefore, from V=IR, the voltage developed across the primary winding is approximately zero when the secondary is shorted. Of course, the secondary voltage is also zero when shorted. But why does this improve the current fidelity? Without going into too much detail, the 'magnetising current' of an xformer depends on the winding inductance and the winding voltage. The inductance is set by the number of turns and the core material and geometry, which are fixed for any given current xformer. However, by reducing the winding voltages, by shorting the secondary, the magnetising current is also reduced to near zero. That is a GOOD THING, because the magnetising current is an error, representing a proportion of the primary current that DOES NOT end up being reflected in the secondary current. The magnetising current results in an error in both the magnitude and phase of the secondary current, definitely not a good thing where high fidelity is required. To say the same thing in a different way, shorting the secondary permits larger currents to be measured with acceptable fidelity, and without core saturation.

The conclusion is a beautiful inverse of the situation for a voltage transformer. In summary:

For a voltage xformer, the primary voltage is fixed, and the current is set by secondary load resistance. Reducing the secondary load resistance increases the current, increases the 'loading', increases the power delivered to the load, and lowers the secondary voltage fidelity. A high value of secondary load resistance is desirable for best voltage fidelity. Shorting the secondary will generate dangerous currents.

For a current xformer, the primary CURRENT is fixed, and the VOLTAGE is set by the secondary load resistance. INCREASING the secondary load resistance increases the VOLTAGE, increases the loading, increases the power delivered to the load, and lowers the secondary CURRENT fidelity. A LOW value of secondary load resistance is desirable for best fidelity. OPEN-CIRCUITING the secondary will generate dangerous VOLTAGES.

Electrical engineering ... truly beautiful.

Colin

> What is the reason, and explain with details.

Most replies answered by pointing out that destructive voltages will be generated if the secondary is open-circuited. While this is true, it does not exactly answer the question.

A current xformer will still operate safely over a wide range of secondary load resistances, so why is a very low secondary load resistance (= approximately a short circuit) the best choice?

The reason is that the fidelity of the current transformer is highest when the secondary is short circuited.

The purpose of a current xformer is to provide a scaled-down version of the primary current with the highest possible fidelity, that is, the secondary current should be a faithful replica of the primary current. In many applications it is not just the magnitude of the measured current that is important, but also faithful reproduction of the phase and high-order harmonics.

Thus, the question becomes, why is the fidelity highest when the secondary is shorted? At first this seems counter-intuitive, for the fidelity of the more familiar voltage transformer is most definitely NOT optimised by shorting the secondary! Indeed, as a voltage transformer is loaded more heavily by reducing the secondary load resistance, the secondary voltage 'sags', phase error is increased, and frequency response decreased, all leading to a loss of fidelity of the secondary voltage waveform. This occurs mainly due to winding resistance and leakage inductance, which would both be zero in an 'ideal' xformer. For a voltage transformer, the fidelity of the secondary voltage is highest with a high secondary load resistance, which draws very little current.

However, with current xformers, just about everything turns out to be reversed compared to voltage xformers, which can be confusing. As with all xformers, the secondary load resistance is reflected back to the primary side, scaled by the square of the turns ratio. Therefore, if we short the secondary, then we also short the primary. Note that the primary current is fixed and constant, being set by the external circuit. Therefore, from V=IR, the voltage developed across the primary winding is approximately zero when the secondary is shorted. Of course, the secondary voltage is also zero when shorted. But why does this improve the current fidelity? Without going into too much detail, the 'magnetising current' of an xformer depends on the winding inductance and the winding voltage. The inductance is set by the number of turns and the core material and geometry, which are fixed for any given current xformer. However, by reducing the winding voltages, by shorting the secondary, the magnetising current is also reduced to near zero. That is a GOOD THING, because the magnetising current is an error, representing a proportion of the primary current that DOES NOT end up being reflected in the secondary current. The magnetising current results in an error in both the magnitude and phase of the secondary current, definitely not a good thing where high fidelity is required. To say the same thing in a different way, shorting the secondary permits larger currents to be measured with acceptable fidelity, and without core saturation.

The conclusion is a beautiful inverse of the situation for a voltage transformer. In summary:

For a voltage xformer, the primary voltage is fixed, and the current is set by secondary load resistance. Reducing the secondary load resistance increases the current, increases the 'loading', increases the power delivered to the load, and lowers the secondary voltage fidelity. A high value of secondary load resistance is desirable for best voltage fidelity. Shorting the secondary will generate dangerous currents.

For a current xformer, the primary CURRENT is fixed, and the VOLTAGE is set by the secondary load resistance. INCREASING the secondary load resistance increases the VOLTAGE, increases the loading, increases the power delivered to the load, and lowers the secondary CURRENT fidelity. A LOW value of secondary load resistance is desirable for best fidelity. OPEN-CIRCUITING the secondary will generate dangerous VOLTAGES.

Electrical engineering ... truly beautiful.

Colin

Hi,

I have a question regarding CT with tap on the secondary.

Lets assume there are 3 terminals on the secondary S1, S2 and S3. Lets also assume that between S1 ans S2 there is a nominal small load (of the protection device, S2 is connected to earth).

What about the windings between S2 and S3?

Should they be shorted in order to prevent dangerous voltage??

I do know that when there is no load on S1-S2, we have to put a short between them. So why S2-S3 shouldn't (since it has no load)!

Thanks,

Shai Halperin

I have a question regarding CT with tap on the secondary.

Lets assume there are 3 terminals on the secondary S1, S2 and S3. Lets also assume that between S1 ans S2 there is a nominal small load (of the protection device, S2 is connected to earth).

What about the windings between S2 and S3?

Should they be shorted in order to prevent dangerous voltage??

I do know that when there is no load on S1-S2, we have to put a short between them. So why S2-S3 shouldn't (since it has no load)!

Thanks,

Shai Halperin

Shai... in answer to your two questions about Multi-Ratio (not Multi-Winding) CTs:

1) Only one tap should be used.

2) Do not short the other.

Regards, Phil Corso

1) Only one tap should be used.

2) Do not short the other.

Regards, Phil Corso

We should connect any one winding to the load or short circuit any one secondary winding, other can be kept open. There will not be any high voltage in the second winding.

Since CT is connected in series with the circuit, we have to make sure that CT secondary circuit is closed through any path and total impedance appearing across CT primary circuit is kept low.

It is just like small resistor (ct)with different tappings connected in series with heavy (load) resister in a circuit, we can use any tap to complete the circuit. if we open entire voltage will appear across the (ct)small resistor.

Thanks

Sunil Kumar (Mangalore, India)

Since CT is connected in series with the circuit, we have to make sure that CT secondary circuit is closed through any path and total impedance appearing across CT primary circuit is kept low.

It is just like small resistor (ct)with different tappings connected in series with heavy (load) resister in a circuit, we can use any tap to complete the circuit. if we open entire voltage will appear across the (ct)small resistor.

Thanks

Sunil Kumar (Mangalore, India)

If it is multi winding (core) then unused winding should be kept shorted because its just like impedance connected in series, if any one secondary winding opens, it became open circuit and cause damage.

Why is not short circuit for ammeter connect secondary of current tranaformer

Thank you for the explanation.

>>Colin et al,

I have a similar question but with a little twist...

What happens to the inductance on the primary side when one of the turns is shorted on the secondary side...ie lets say you have 10 turns on primary and 100 turns on secondary and you solder turn 50 and 51 together (not to ground or anything else just each other). What happens?

I have a similar question but with a little twist...

What happens to the inductance on the primary side when one of the turns is shorted on the secondary side...ie lets say you have 10 turns on primary and 100 turns on secondary and you solder turn 50 and 51 together (not to ground or anything else just each other). What happens?

Ali Ali... very good question. Essentially there are two 'negative' results for the example you cited:

1) The turns-ratio Nsec/Npri will be reduced by one turn, that is, from the normal 10:1 to 9:1. This of course will affect measurement accuracy!

2) More importantly, transformer-action will magnetically couple the 'shorted-turn' to the other 99 turns having and effective turns-ratio of 99:1. The resultant induced current in the single loop will produce localized heating that will cause additional damage to insulation, winding, dand core!

The above discussion ignores other exacerbating factors such as frequency, current wave-shape (if not sinusoidal) and capacitance, if present.

Regards, Phil Corso

1) The turns-ratio Nsec/Npri will be reduced by one turn, that is, from the normal 10:1 to 9:1. This of course will affect measurement accuracy!

2) More importantly, transformer-action will magnetically couple the 'shorted-turn' to the other 99 turns having and effective turns-ratio of 99:1. The resultant induced current in the single loop will produce localized heating that will cause additional damage to insulation, winding, dand core!

The above discussion ignores other exacerbating factors such as frequency, current wave-shape (if not sinusoidal) and capacitance, if present.

Regards, Phil Corso

Ali Ali... correction to my earlier comment:

1) Nsec/Npri ratio for shorted turn-to-turn condition should be 9.9:1!

Phil

1) Nsec/Npri ratio for shorted turn-to-turn condition should be 9.9:1!

Phil

All comments where very helpful. In addition, i think one can also use the transformation formula V1/V2=I2/I1 to quickly asses the secondary voltage V2 which is equal to V1xI1/I2. The answer is very obvious: if I2 is zero, the result is infinity. thank you.

With this, how do you explain the fact that the secondary of a PT should always be kept open? :)

> All comments where very helpful. In addition, i think one can also use the

> transformation formula V1/V2=I2/I1 to quickly asses the secondary voltage V2

> which is equal to V1xI1/I2. The answer is very obvious: if I2 is zero, the result is infinity.

> All comments where very helpful. In addition, i think one can also use the

> transformation formula V1/V2=I2/I1 to quickly asses the secondary voltage V2

> which is equal to V1xI1/I2. The answer is very obvious: if I2 is zero, the result is infinity.

Though the secondary voltage is very very high, it is not infinity and there is a correct way to calculate it. Thanks.

> All comments where very helpful. In addition, i think one can also use the transformation formula V1/V2=I2/I1 to

> quickly asses the secondary voltage V2 which is equal to V1xI1/I2. The answer is very obvious: if I2 is zero, the

> result is infinity. thank you.

> All comments where very helpful. In addition, i think one can also use the transformation formula V1/V2=I2/I1 to

> quickly asses the secondary voltage V2 which is equal to V1xI1/I2. The answer is very obvious: if I2 is zero, the

> result is infinity. thank you.

> The current transformer secondary is short circuited during the operation.

ANS:. If the transformer secondary is not short circuited, during the operation - it will opened means, it damage the coil and burnt out. It will affect the surrounding peoples also. It is very dangerous.

ANS:. If the transformer secondary is not short circuited, during the operation - it will opened means, it damage the coil and burnt out. It will affect the surrounding peoples also. It is very dangerous.

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