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Open Circuit a Current Transformer
Why should you never open circuit a CT?

I know you should never open circuit a CT because it will lead to massive voltages being developed in the secondary and possibly fire. But why does it do this? Why doesn't a power transformer destroy itself if you open circuit it? What makes the 2 different? Any information I would be hugely greatful for.

Many Thanks.
Josh Banks

By Phil Corso, PE on 11 March, 2006 - 3:56 pm
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Responding to Josh's Mar 10, 5:45pm query... without getting too technical:

Q1) Why does opening a CT's secondary cause dangerously high secondary voltage?

A1) With the flow of both primary and secondary currents the transformer's exciting current is very low. The secondary current mmf serves to keep the magnetizing flux in check! If the secondary now opens the primary current mmf produces an exciting current 'orders of magnitude' greater than normal. And, the resultant large increase in flux density produces an extremely high voltage in the secondary.

Q2) (Paraphrasing ) Why don't power transformers exhibit the same phenomena?

A2) While current transformer (CT) and power transformer (PwrT) theory is about the same the salient difference is their Volt-Ampere characteristic. The PwrT is a constant-voltage (shunt-loaded) device whose primary and secondary VAs are equal. Conversely, the CT is constant-current (series-loaded) device. That is, the input VA is insignificant, having no influence upon the primary circuit. Its inter-related variables like flux, volts, current, and burden are not constant but more likely non-linear.

If you need more technical detail than given above, contact me!

Regards, Phil Corso, PE {Boca Raton, FL, USA} [] (

why exciting current is low when primary and secondary is on?

1 out of 1 members thought this post was helpful...


I suggest you search Control.Com Archives using the term +"Current Trans", in the SEARCH THE SITE box. There are some 35 CT threads!

Phil Corso

By anilreddy488 on 3 April, 2017 - 2:42 am

Mr Corso,

Could you please further elaborate on the voltampere characteristics of a Power Transformer and Current Transformer

Anil Reddy, DE,

By Anonymous on 11 March, 2006 - 3:57 pm

High voltages, see this article

By Curt Wuollet on 12 March, 2006 - 12:00 pm

Simply transformer action. The turns ratios of most power transformers are low so the voltages are low also. In a CT the primary is 1 turn so the turns ratio is high. Normally the primary voltage drop is low because of the reflected impedance of the load which is usually a very low resistance like a current shunt. When you remove the load, the primary drops appreciable voltage according to it's now higher inductive reactance and the voltage is multiplied by the high turns ratio and you have sometimes very high voltages. The biggest danger is when you open the load under power as the voltage rises quickly enough to maintain an arc. If the transformer or something connected to it is not insulated for high voltages you can get a fire. If Jumper Joe is pulling the wire off, he can become the easiest path to ground.


1 out of 1 members thought this post was helpful...

Responding to Curt's Mar12, 12:00pm and Meir's Mar 12, 12:02pm comments:

Tsk, tsk... I too appreciate not-to-technical answers... but, they should be technically correct:

A) Turns-ratio is not responsible for overvoltage. Even CTs with lower ratios, like 5:1 or 1:1, have the cautionary notice (at least in the USA!) Furthermore primary voltage is a fraction of Volts. So, using turns-ratio as the primary to secondary multiplier, secondary volts would be in the order of volts to tens of volts.

B) Overvoltage occurs when the secondary open-circuits and the demagnetizing effect of the secondary emf is lost. Flux density quickly increases, limited only by core saturation.

C) As core saturation progresses, waveshape changes. The usual sinewave changes to a peaked wave. Such a waveshape has an extremely high dV/dT characteristic. It is very likely that insulation flashover and/or fire results from dV/dT... not from a high amplitude sine wave!

Phil Corso, PE {Boca Raton, FL, USA}
[] (

By Curt Wuollet on 18 March, 2006 - 10:57 am

Yes, but it is technically correct. You are also correct in that there is di/dt action going on. The theoretically instantanious collapse of the field is going to cause theoretically infinite voltage. And that field is generated by primary current. You will find that within physical limits, the change in flux density affects both windings. Actually a better model than "normal" transformer action might be the common Kettering ignition coil. And I don't recall stating that the voltage would be sinusoidal although you might get lucky and interrupt the sinusoidal current at a zero crossing. In any case, it's a bad idea. The non-technical explanation is you are going to get a big spike to light an arc and greater than normal voltage to help keep it going.


By Meir Saggie on 12 March, 2006 - 12:02 pm

Basic electrical theory (which I am not going to go into) - an open CT present itself as a large inductance to the primary current source and develops a large voltage on it primary winding. The secondary winding will then develop a voltage that is N times larger than the primary (because the transformer ratio is 1:N - Primary:Secondary).

By Gary Wiggins on 17 March, 2006 - 11:49 am

The CT is designed to lower current to a safe and measureable level. To accomplish this, the voltage level on the secondary is raised. P=IE, if P is constant, I and E are then inversely proportional. The windings ratio determines the secondary voltage levels. While the secondary is connected to a measuring device, it is essentially shorted. Opening the secondary circuit immediately raises the voltage to levels pre-determined by the primary voltage and the windings ratio. There are shunted CT's available that greatly reduce potential hazards typically associated with standard design CT's. Always remove power and install shorting bars prior to working on CT circuits.

By Shailesh Patel on 18 March, 2006 - 10:59 am

Dear josh Case : Power through aprox. transformer is constant

when you think about power transformer then voltage is aprox. constant and current in primary is depend on secondary load. so if you short power transformer then current in primary become very huge and may (must) be damage to winding and insulation.

But about currant transformer current passing in primary is depend (=same) on busbas or cable where current transformer mount (which is not dependent on load of secondary of CT) & voltage of primary is depend on secondary load. if your secondary of CT is open than as per magnetic coupling very high voltage generated (because of secondary current = 0 ) at secondary. And insulation will be fail.

Shailesh Patel

1 out of 1 members thought this post was helpful...

A few off-list questions have prompted revisiting this topic, as well as two related ones, namely, "Current Transformer" (#1026172605) and "Current Transformer Failures" (#1026113301). From the questions asked it is obvious the topic is still in a state of flux (pun intended!)

Some commenters expressed a 'very strong' belief that overvoltage is due to CT turns-ratio. Others, myself included, believe that while ratio is present, it is not the reason for the high voltage. A new approach is presented here to resolve the issue.

Consider the example submitted by Pagnair... a CT whose current ratings are is 2,000/1/5. But, ratio alone does not complete the specification. So, for calculation purposes, I will assume it's capacity is 10 VA (volt-amperes) for the 1-Amp tap.

Tagging primary and secondary parameters with subscripts p and s, respectively, letting the winding turns-ratio = Np / Ns, then:

A) Primary Parameters. The pri / sec current-ratio (Ap / As) is 2,000:1. Primary voltage, Vp, is determined by the simple calculation VA / Ap, or 10 / 2,000 = 5.0 mV. Primary ampere-turns, ATp, are 2,000 x 1.0 = 2,000 AT.

B) Secondary Parameters. Secondary voltage, Vs, is VA / As = 10 Volts. Secondary ampere-turns, ATs, are As x Ns / Np = 2,000 AT!

C) Ampere-Turns Impact on Core-Flux. Note, from above, the primary and secondary ampere-turns are equal. The direction of their flux flows are in opposition. But, a small core-flux is still required to cover both iron-core and winding losses. This net core-flux, ATe, is quite small, as low as 1/2% to 2% of the rated ampere-turns. (An aside, it is this value that establishes the CT's accuracy!)

D) Net Core-Flux Produces Output. Using a 1% value, the exciting ATe is, in this case, 20-AT. This inturn (pun not intended) results in an output voltage of 0.5 Volt per ampere-turn.

E) Effect of Opening Secondary. Opening the secondary results in the full primary ampere-turns increasing the core-flux, i.e., from 20 to 2,000 ampere-turns. This of course produces 2,000 AT x 0.5 V / AT volts, or 1,000 Volts at the secondary.

F) Conclusion. CT-ratio is not the cause of overvoltage. Instead it is the substitution of an overwhelming primary ampere-turns for the very small excitation ampere-turns! Note, the calculations above are based solely on an rms-treatment of both flux and voltage. Core-saturation effects were not considered!

Regards, Phil Corso, PE {Boca Raton, FL, USA} [] (

By Phil Corso, PE on 16 September, 2006 - 7:48 pm
1 out of 1 members thought this post was helpful...

You know I'm a stickler (some say obsessed) for Technical Correctness. Several contributors to this and related current transformer topics made
several errors:

1) When transferring the secondary impedance, Zs, into the primary as Zs', the correct expression is,

Zs' = Zs x (Np/Ns)^2, where,

Np and Ns are the primary and secondary winding turns, respectively.

For the CT example cited the secondary impedance reflected into the primary is,

Zs' = 10 x (1/2,000)^2 = 2.5E-6.

2) This low value illustrates that secondary impedance has no influence on the primary circuit... contrary to what some stated.

3) Some presented the formula for the development of an extreme secondary voltage as Es = Ep x current-ratio! Again using the CT cited, for an open circuit condition the secondary voltage would have been 5 mV x 2,000 or 10 Volts. Clearly not the extreme voltage observed!

If any of you disagree, then present a point of view free of obfuscation!

Phil Corso, PE {Boca Raton, FL, USA}
[] (

By Sten Maraldo on 19 December, 2006 - 6:11 pm

I have read with great interest the various comments expressed and I feel that a number of questionable interpretations have been presented.
Under open circuit conditions the total primary current will be equal to the total excitation
force of Ampereturns/metre, thus inducing maximum
induction (B of 1.9-2.0 tesla for normal materials).

The voltage generated in the secondary winding under open circuit will be directly proportional to:
Number of secondary turns
Cross sectional area core

Extremely large voltages can be generated in Cts that have large turn ratios, i.e. 5000/1, and have
large CSA as typically used in protection class cts for differential protection.

My experience in laboratory tests shows that in CT designs with AT/m up to 1500, the relationship of generated voltage is linear and proportional to the items listed above. I have developed curves to estimate open circuit voltages and would greatly appreciate to have some comparisons if anyone has similar information.

Sten Maraldo

By Darryl Muetzel on 24 March, 2007 - 6:23 pm

Mr Corso,
I have a problem - maybe you can point me in the right direction.

I have an application where my customer is requiring FS(instrument security factor)>=5.

We were unable to meet this with the BCT design - so we suggested to use a secondary protector (current shunt) that would short circuit when the voltage across the burden would reach a certain voltage (due to over current during a short circuit event).

This was agreed to - but when I went to the vendor (ITI/GE) to see if they could make up a secondary protector (similar to their OCP line of open circuit protectors) with a lower activation voltage than their current designs- they said that they are unable to design any new devices for several reasons - which aren't really important.

SOOO - for my question - do you have any information on who could manufacture such a secondary protector (activation voltage is 60V)?

Any help you (or anyone else) could give would be very much appreciated.

Thanks - Darryl Muetzel (

By Phil Corso, PE on 25 March, 2007 - 11:52 am
1 out of 1 members thought this post was helpful...

Darryl, cts have been used in protective relaying applications for about a century. Your attempt to use a new technology has me puzzled. Please provide additional information:

1) What are details (primary kV, primary Amp, ratio, SC level, burden impedance, etc)?
2) Why do you believe an overvoltage will occur?
3) Are you concerned about saturation effects?
4) Why the intent to use a new technology?
5) What is the customer's reason for Inst FS?
6) What, if any, is the cost-to-benefit ratio?

Phil Corso, PE (


Q1) What will happen when CT secondary is kept open and energize the primary?

Q2) Is it only dv/dt or di/dt relation for overvoltage across CT sry winding?

Open Circuit is dangerous only with shortcut current. Nominal current is not so high level.

By Manu Thomas on 13 April, 2007 - 12:54 am

The primary current in the ct is not controlled.
This is determined by the load on the power side.
When a primary current flows, a secondary current is indused as per the ct ratio. If we keep the ct secondary open, it tries to develop a voltage to pass that secondary current. This cause insulation to break down in ct.

Manu Thomas

I think that when we open ct and there is current running in the primary coil this cause high voltage in secondary coil but pwr tr when is open no current running in primary coil
hozgab [at]

1 out of 1 members thought this post was helpful...

we made an experiment.
A 3000/1 Current transformer was gradually energized in steps with secondary open right from beginning.

We observed that at 10% rated current in primary. after about 10 seconds, the CT failed. The secondary when tested was, found to have shorted, also confirmed by significant reduction of winding resistance indicating interlayer shorts. Since we did not open secondary while primary was energised, it is not a Ldi/dt voltage failure.

Also, when tested earlier, this CT showed a knee point voltage of 750 volts at the secondary, which is well below the insulation level of the enamel wire used for secondary, i.e. 2 kV.
Thus even when open, the voltage developed may have exceeded the knee voltage to a saturation level of the order of at most 1 kV.

Thus I am not clear. what c`d be the cause of insulation failure?

2 out of 2 members thought this post was helpful...

Dhamdev... good work!

Regarding your question about not reaching the varnish breakdown voltage when you tested for the kneepoint, the CT had not yet been saturated.

Regards, Phil Corso

2 out of 2 members thought this post was helpful...

Dhamdev... Further to my earlier comment, the application of 10% primary current the applied ampere-turns (AT) is about 30-45 times the nominal exciting-current AT.

Phil Corso

To get the answer for this let us understand the working principal of power transformer first. If you check the power transformer, when you apply voltage initially currents are very high (Magnetising current), but as soon as core get magnetised counter EMF induced in secondary winding which will oppose the primary EMF.

In power transformer counter EMF in secondary winding is playing role in controlling primary current and complete supply voltage to appear across primary winding and power transformers are designed for Higher VA rating.

Let us imagine the same power transformer without Core and secondary winding, now if you energise the transformer it will become short circuit and short circuit current will flow through primary winding, why because there is no Counter EMF induced in secondary winding which can oppose primary.

If you talk about CT it works same like transformer but it has designed with very small VA rating hence it cannot control primary current because of weaker counter EMF induced in secondary winding. and if you keep open secondary winding there is no current to flow and voltage will rise (Counter EMF) which will try to counter primary EMF but its is weaker as per design hence it will end with insulation failure or some time CT may blast.

As per my knowledge the basic working principle of Power transformer and C.T are the same, but they are designed as per the applications.
This is just my opinion, it may not be correct always welcome for the discussion which can bring better result.

Thank you
Sunil Kumar

By Colin Dedman on 19 October, 2011 - 1:56 am

Conventional xformer theory explains and fully predicts why large secondary voltages are generated if the secondary winding of a current xformer is open circuited.

A good place to start is to consider the properties of an 'ideal' xformer, noting that there is no fundamental difference in the operating theory of voltage and current xformers.

As explained in any textbook, an ideal xformer has infinitely high winding inductance, zero winding resistance, perfect magnetic coupling between windings, and 100% efficiency. The impedance transformation ratio is equal to the square of the turns ratio. At all times the primary amp-turns is equal and opposite to the secondary amp-turns, so that the net amp-turns is always zero. This is all true for a 'perfect' xformer, and a high quality real xformer will closely approximate this behavior.

In an ideal voltage xformer, we apply a fixed voltage to the primary winding, and if the secondary is unloaded, then the primary current is zero because the primary inductance is infinite. No nasty surprises, unless we short-circuit the secondary, resulting in very large winding currents.

Now let's look at so-called current xformers, noting that an xformer is an xformer, and all of the ideal properties are identical. Normally with a current xformer, a resistive load is connected across the secondary, and this resistance is 'seen' on the primary side, multiplied by the square of the turns ratio. For example, if Nsec/Npri=100, and the secondary is loaded with 10 ohm, then the resistive load seen on the primary side is 10/10000 = 0.001 ohms. In effect, the primary current is in series with a 1 milliohm shunt resistor. If the primary current is 100 amps, then the voltage developed across this primary 'shunt' resistance is 100x0.001=0.1 Volts. This primary voltage is stepped up by the turns ratio, so the secondary voltage is 100x0.1 = 10 volts. As a check, note that the primary and secondary powers are equal, ie, 100A*0.1V = 1A*10V. You can also check that the primary and secondary amp-turns are equal. (They are) So far, no problems, and all is easily understood in terms of well-known xformer theory.

NOW CONSIDER WHAT HAPPENS IF THE SECONDARY LOAD RESISTANCE IS INCREASED. Instead of Rsec=10 ohms, what if Rsec=1000 ohms?? The primary current is still 100A, because that is set by the external circuit, not by the current xformer. In our example, with Rsec=1000 ohms, Rpri=0.1ohms, Vpri=10V and Vsec=1000V. Hmmm. Clearly, for an ideal current xformer, as the secondary load resistance is increased towards infinity (=disconnecting the secondary load), the primary and secondary voltages will rise towards infinity. Real xformers approximate ideal xformers quite well so that, my friends, is why you should never open-circuit the secondary of a current xformer.

There is a clear analogy here with voltage transformers. With a voltage xformer, where a constant voltage is connected across the primary, never short circuit the secondary, or dangerously large currents will be generated. With a current xformer, where a constant current flows through the primary, never open-circuit the secondary, or dangerously large voltages will be generated.

Hope this helps.


This is the most concise, clear, explanation that I have read. It makes perfect sense to me. Only having what I consider basic electronics knowledge, this explanation has gone a long way for me. I kept saying .... but they are both just can a potential transformer be so predicable, and a current transformer be a magical box? Thanks for clearing that up. From a theory standpoint, they really are the same ... imagine that! Just matters whether the supply is constant voltage (like the ones we use all the times in our daily lives) or constant current - I guess that could even be my little clamp on amprobe meter?


To better understanding of CT let us connect two single phase power transformers TrA & TrB, in series which are equal rating (for example 11kv/33kv, 1000/100Aamps and %impedance is equal) in series to 11kv supply.

Now let us consider both transformer are equally loaded, then voltage accross each transformer should be half of supply voltage;5.5kv aprox (equal impedance)and current is equal at both transformer because connected in series.

Now let us consider both transformers at different load, now the voltage drop across each transformer is depends upon their respective circuit impedance but current through both transformers are same and depends on total impedance of the circuit I=V/ZTrA+ZTrB(series circuit).

Now let us open circuit one transformer then the impedance of that transformer become infinity and the voltage acrross that transformer will rise upto the rated voltage(Supply voltage 11kv/33kv) but here transformer has designed to operate at rated voltage, hence there will not be any insulation failure.

Now let us consider CT inplace of one of two trasformers, Now when the CT is connected to certain load. the impedance of CT is negligible compare to the power transformer connected in series. Hence voltage drop accross CT is negligible. Current in the circuit is I=V/ZTrA+Zct( connected in series)Here Impedance of Current transformer is negligible hence the load current is totally depends on ZTrA, ie.I=V/ZTrA.

Now let us consider CT open circuited. Now CT Impedance become high and it will try to increase its voltage, but far before insulation fails, because it has desinged to operate at certain load or with short circuit condition.

Hope this explanaiton may help someone.

Thank you
Sunil Kumar (Mangalore, India)

All of the above comments explains what happens to the voltage when a ct circuit is opened, but what happens to the current? Doesn't the current need to somehow find a path back to the ct it came from? Does it just circulate inside the winding of the ct? I don't have a handle on what happens to the current.

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When current flows in the primary of a CT with an open-circuited secondary, the first thing that happens is it hums--very loudly. And the second thing that happens is that it basically explodes and any flammable components burn. This doesn't take long, and the time is inversely proportional to the magnitude of the primary voltage--the higher the voltage the faster the explosion/fire occurs.

I've heard CTs with open-circuited secondaries hum very loudly when the primary was energized but no current was flowing, and the breaker that was closed to energize the primary is quickly opened, manually. I've never seen on energized for more than a couple of seconds without current flowing in the primary, but they don't last that long--even at 440 or 480 VAC primary voltage when current is flowing in the primary.

So, the current doesn't do anything for very long. First of all, it never had anywhere to go if the secondary was open-circuited. Second, when current does flow (usually at the secondary terminals of the CT which are very close together), there is a large spark and near simultaneous explosion.

That's why most CT secondary circuits have shorting means that are supposed to be inserted in the circuit before the secondary circuit is ever opened for any reason. And the shorting means are usually located very close to the CTs themselves.

The upshot of all of this is: Don't ever energize the primary of a CT if the secondary is open-circuited. Someone will be, at least, be buying and installing a new CT, and possibly cleaning up a large mess and replacing some wiring, maybe even the primary wiring/bus bar.

> I know you should never open circuit a CT because it will lead to massive voltages being developed in the
> secondary and possibly fire. But why does it do this? Why doesn't a power transformer destroy itself if you open
> circuit it? What makes the 2 different? Any information I would be hugely grateful for.

Part 1 of your question:
A Current transformer designed to withstand open circuit would be more expensive, and may have worse characteristics in normal operation.

Explanation - (Warning: Technical)

What is a CT?
Usually a current transformer consists of a single 'turn' of wire through an iron core for the primary, and a few hundred turns for the secondary. If you understand what an idealised component model is, you can think of it as an inductor in parallel with an ideal transformer.

The voltage drop across the wire 'primary' in an unloaded CT will be V=L*(di/dt).
Where the inductance L is determined by properties of the wire and the core of the CT, and (di/dt) is change in current with respect to time.

This voltage will, by transformer action occur at the secondary. This unloaded voltage will be proportional to the current, frequency and turns ratio. The constant of proportionality depends on the core material and the wire used to form the 'primary'. As such, the open circuit method a not very useful way to measure current.

In normal operation a low impedance is placed on the secondary, this appears as a very low impedance across the primary - much lower than that of the 'inductor'. The output current will now be (mostly) directly proportional to the input current and the turns ratio.

Part 2:
Simply, a power transformer is designed to withstand the full supply voltage, while a current transformer is designed for a tiny fraction of that supply voltage.

If the current transformer were of ideal construction, and the secondary went open, then the infinite secondary impedance would be reflected into the primary. The 'primary' would then appear open circuit and be subject to the full supply voltage.

Source: Electrical Power Engineering lecture notes - Current Transformers.

> I know you should never open circuit a CT because it will lead to massive voltages
> being developed in the secondary and possibly fire. But why does it do this?

I think about a CT in simple transformer turns e.g. you have a 500:5
so that's 100 turns on the secondary for 1 turn on the primary.
Now suppose the CT is measuring the incoming buss on a 400 Volt switchgear and the secondary becomes open, the impedance of the transformer shoots up and with a substantial load on the switchgear it tries to drop the whole 400 Volts across the CT, that's 400x100 across the secondary or 40 kV.

I know this is not strictly correct but it explains it well enough for me.

Many years ago I was given the task of checking the overload relays on some medium Voltage switchgear, the engineer insisted we inject the equipment right on the busbars, we did this with a 1 kVa transformer with about 4 parallel turns of welding cable for a secondary, it would easily put out 2,000 Amps. On injecting one set off gear we couldn't get more than a few hundred Amps, we found there was a wiring mistake on the CT secondary side and the CT impedance was limiting the current. This would have gone un-noticed using the usual relay testing equipment.

In all comments there's no mention of core saturation.

The field strength has limited value. So, while the waveform will be messed up the CT open circuit voltage has limits governed by the size of core.

the core is rated for so many ampere-turns without saturation. you need secondary current to get saturation effects. basically saturation under an open-circuit secondary is not likely, until secondary arcing take place.

ferro-resonant failures are unlikely with current transformer installations. that only occurs in large sub-station transformers under low loads, and with power correction caps in the primary that have not been switched out of service.

I've been reading this thread with some interest because the danger of an open circuit CT figures largely where I work. I've puzzled it over and come up with this explanation. Hope it makes sense:- When a CT is loaded the primary and secondary ampere turns are balanced. A small amount of primary current (~1%) provides CT excitation. If the secondary is opened, all the primary current now drives the excitation. The flux rises to approx. 2 tesla and then the flux waveform flattens as the core becomes saturated. The flux waveform has changed from a smooth sinusoidal to more of a square wave where the rate of rise and fall is roughly the same as the primary current about it's zero crossover. So, if we take the gradient of the primary current at it's zero crossover we get dI/dt=(Imax)X(2xpixf). This steep rise and fall corresponds to a large dflux/dt at the start and end of the 'square' flux waveform. The secondary voltage is Vs=-Nxdflux/dt. So,for a 1000:1 CT, using the Primary current's crossover gradient and 1000 turns, the secondary voltage is roughly -1000x(Imax)x(2xpix50) which is 314200xImax. If Imax is 100 amps, we get a very large secondary voltage.

Basically, the voltage developed in a coil, from a fundamental point of view, is directly related to the rate of change of flux. So, a high rate of change gives a high voltage.

With a CT secondary open circuit, and the CT excitation driven by the Primary current, the rate of change of flux is very high at the zero crossover. The core then becomes saturated for most of the sinusoidal half cycle. The voltage output will be seen as spikes at the zero crossovers.