Power System Protection

Professor A K Pradhan
Department of Electrical Engineering
Indian Institute of Technology, Kharagpur
Lecture 33
Current Transformer – Part - 1
(Refer Slide Time: 00:33)

Welcome to NPTEL course on power system protection. We will start with the current and voltage
transformers. In this lecture we will discuss on current transformer, the equivalent circuit of the
current transformer, the burden to the current transformer calculation and the classification aspects
on current transformers.
(Refer Slide Time: 00:57)

To the relays the voltage and current signals are the inputs and these inputs as we have already
mentioned are from sensors, current signals from current transformer, and voltage signals from
voltage transformers. The whole objective of these sensors or transducers are to lower the
magnitude to scale down the corresponding signals to a suitable level which is compatible to the
relays and to provide galvanic isolation, electric isolation between the power network which is a
very high voltage and the relays and the other connected instruments in the system. So, we will
see that earlier in all our discussions on relays you assume that the sensors provide the signals of
the systems perfectly and based on that considerations we see that.

Current transformers, secondary windings are rated 5A or 1A different countries used consider
different perspective different system requirement may be different also an agency may be using
in general either 1 ampere or 5 ampere, voltage transformer some countries used 110 V some 120
V phase to phase voltage or in terms of phase to neutral it is 63.5 or 69.3 V hold respectively.

The point here is that this higher voltage of 110 V is not compatible to numerical relay, because
we are confined to low voltage like 5 V or 10 V. So, the practice of this CT and VT’s are pretty
old and they are being used when electromechanical relays are there in the high voltage system
also. So, to have a retrofitting perspective the corresponding CT’s and the VT’s are still continuing.

So, in that respect the numerical relay may be using subsequent scaling down the signals
incompatible to its own internal arrangement. Furthermore, we know that the system like during
fault see very large amount of current, so this current has to be scaled down by the CT’s, therefore
the CT primary we will see particularly during fault very larger current.

That means that the corresponding current may be as high as 50 times of the rated current, so
corresponding CT has to manage that such a large value without any possible damage to it and that
maybe for few seconds also because the fall may persist for that kind of period. On the other end
system voltage may be sometimes higher during different dynamic condition in the systems;
therefore the corresponding voltage sensors VT typically considered having 20% above the normal
value to withstand for a very long period of time.

These sensors current transformer, voltage transformer subsequently discuss also capacitor
coupling voltage transformer or simply capacitor voltage transformer, they are required to scale
down the voltage to a lower level. Now, point is a voltage transformer and the current transformer
in general is associated with a magnetic circuit, so that the galvanic isolation is being provided.

Any magnetic circuit if it goes to the saturation, then the corresponding transformer current ratio
or the corresponding terminal voltage that becomes different one. In that perspective for a current
transformer the current signal may be very large at times and if it reaches to the saturation region
of the current transformer.

Then the information on the primary side which will be conveyed to the relay using the secondary
of the current transformer may not be the correct you can say that corresponding correct value of
the primary. That puts challenges and that is the reason we have this lecture on current transformers
and voltage transformer.
(Refer Slide Time: 07:08)

We will continue with current transformer first and then we can say that in subsequent lectures we
will see the voltage aspects. Coming to the current transformer; simply if this is the conductor in
the high voltage system and we have a core and we have a winding, so this is the secondary winding
and this is the primary of the current transformer.

So, you see here that essentially require large number of turns in the secondary, the reason behind
that the current has to be reduced. Therefore the number of secondary turns will be accordingly
should be as many. Now, there is another type of current transformer you will see in industry that
is called flux summing CT.

So, what happens that in the gap of the corresponding core a toroid or so as it seems, then all the
three conductors go through with proper insulation and because the this area available is smaller,
that is why this is more suitable for only for low voltage system and this you can say that if the
corresponding number of turns for the secondary and then we have to consider the corresponding
connection to the relay and so and this relay you can say that different winding impedance Zb
including the corresponding connecting wires and so.

In this case the corresponding summing CT so all the currents Ia, Ib, Ic will be added which is
nothing but gives us thing 3I0 or the corresponding residual current and so. In this case if we see
from this difference all the current will be added, therefore the CT ratio is independent over the
load current; so individual load current in all three will be added and that corresponds to that 3I0.
So, normally if it is a balanced systems that becomes 0 and during fault that unbalance amount of
zero sequence current will be sensed by the secondary.

Therefore, it is the independent of CT ratio, load current. Whereas here if we think about the
corresponding CT and the individual phase current associated in this arrangement and that's why
it is associate also with load current perspective while thinking of the number of turns.

Note that such current transformers which are being used in the industry are magnetically coupled
that we have already mentioned, so that is why need a core and they are having multi winding in
transformers, there are multiple windings in this system. The high-voltage current transform may
content also several cores, not single, they can have multiple cores and each with a secondary
winding individually.

So, that you can say that each corresponding windings can be used for protection can be used also
for control and also for metering, individual winding can be used for different applications. For
example, in a substation in Indian grid it is found that there are 5 cores that the CT have, so 1 core
is connected to the distance and differential relay, core 2 winding is connected to the further
distance and differential relay. Core 3 is connected to metering, core 4 is connected the bus-bar
differential and core 5 is connected to the bus-bar differential perspective. So, what you can say
that all the cores and the associated winding are enclosed in one unit. That is what we can talk
about multi winding and also we can say that multi-core CT’s.

(Refer Slide Time: 11:23)

Current transfer ratio that I1 : I2 and 1:N, because there are number of turns you can say that in the
secondary will be higher, so IEEE standard says you can say that the 600 : 5 you can say that multi
ratio current transformer, now it is multi ratio. So, what is being done we can say that that from
the 600 : 5 so this is 1 : 120, that means that the number of turns available in this secondary from
the different number of turns you can have a tapings and then accordingly the ratio will also be
different.

So, that you can say that tapings and the corresponding ratio can be used for different CT
transformation, like for this 600 : 5 the manufacturer should provide the standard ratio of 50 : 5,
100 : 5 and continuing like this at the end 600:5. So, if we say from this 600:5 CT gives you 1:120
the turn’s ratio, then 50: 5 it is 1: 10 turn ratio. From these different tapings you can have different
current ratio available from the systems.

Other available CT’s you can see that similar things are available as per the standard on 1200: 5,
2000: 5, 3000 : 5, like that 5000 : 5 and so. On the selection of the CT’s the CT ratios that are
selected to meet maximum load current requirement, because this is a continuous rating, faults are
rare in the system, primary of the CT should meet the maximum load current, that is what we saw
this also in the distribution system overcurrent relay design also.

CT ratio should be large enough, so that CT secondary current does not exceed 20 times the rated
current under the maximum symmetrical primary fault current. So, we say that CT ratio should be
large enough so that the CT secondary current does not exceed 20 times. It means that for fault
conditions the CT secondary current should not go far beyond 20 times above its rating of this one
that is what we can say that a guiding factor for this. Accordingly you can say that the CT ratio
should be selected for a given system.
(Refer Slide Time: 14:06)

On furthering the current transformer ratio we say you can say that here one example let us say
you can say that a CT lead runs 200 m in high voltage substation, so that resistance of you can
say that the where the CT lead can be as high as you can see that around 3 Ω, Now, let us have
simple calculations in terms of the burden, in CT burden we will define in terms of the burden if
this Zb you can say that is talk about the burden prospective, so then which is the corresponding
wire impedance in the secondary and also the corresponding relay or any other you can say that
the elements connected to the secondary of the CT’s.

So, that impedance if you see Zb then the burden to be consider transformer its I2Zb, so that is the
load you can say that connected in the secondary load to the secondary winding, so that is what
called as

CT Burden VA = 𝐼 2 𝑍𝑏

Zb is the equivalent impedance of the connecting lead and the elements connected to the secondary
winding. Let us, consider a 5A CT and we have 3Ω lead as in this system, so that leads to a burden
amount of I2× 3, so this is continuous rating 5 A, 75VA, then we have a relay burden, let us say
electromagnetic relay typical burden of 10 VA and for numerical relay maybe less than much
lesser than 1 VA. So, considering the higher side (75 + 10) VA gives us 85 VA, but this 85 VA
for the CT seems to be very high and accordingly you can say that because VA is more so the
corresponding column so that the conductor size of receiver we consider will be thicker and which
will remains pretty expensive. If you go to the 1 A CT the corresponding current rating is small,
so 12 ×3becomes 3 VA plus the burden of the electromagnetic relay assuming 10, 10 + 3 becomes
13. So, as compared to 85 this becomes 13, so we can see say the see here that the burden is much
smaller, so the expected you can see that the economic aspects should be smaller.

Only you can say that additional thing here is that the number of turns in case of 1 A will be a
more you can say that that is compare to the 5 A CT. So, we can have a meaningful comparison
of these corresponding size, weight and cost between these two and then someone and before this
but those are you can say that again constants in turns to the utility what they have the common
practice and so.

So on the basis of that preference technical difference between 5 and 1 A kinds of thing. Now, in
case of very high that connecting in the high voltage system also, 2000 A and so we can say that
the secondary current can be used to limit the number of secondary turns, sometimes you can say
that we may prefer a 5 A over 1 A because of the reduced number of turns.

Sometimes in exceptional case you can go for 20 A CT, where you can use another level of
transformation with an interposing CT that of 20: 1 also. So, this is about something on the current
transformer ratio selection.

(Refer Slide Time: 18:03)

One thing that we see that for a 3-ɸ systems each phase you can see that current has to be taped
and that should be the input to the relay, so the corresponding currents are being process in the
relay and always there is a polarity of the corresponding current, then we cannot integrate the
corresponding currents properly. Therefore, while connecting the CT’s we essentially require the
corresponding polarity of the transformer. CT or VT or so you can that any transformer with the
corresponding polarity marking, so conventionally what is being done which gives us the
corresponding direction of the winding in the core.

So, we put a solid mark at the both the high voltage and the low voltage or in the secondary side
with this starting points, so we can say that this mark and this mark at the starting point, what it
shows that with respect to this starting point then if you go to the other end, then in both that
windings in the core will be wound in the same you can say that sense, either clockwise or
anticlockwise.

So, that you can say that this is being reveal by this marking. In some of the cases you can say that
if it is being mark you can say that also H1, H2, high voltage side, X1, X2 in the low voltage side
also, so and so in the transformer also. And this helps is you can say if the current is entering to
the marked terminal the solid mark or the H1 and the current you can say that I which are leaving
the mark terminal then both the currents will be almost in phase and the corresponding voltage
from the other terminal to that mark one in this case also, these V1 and V2 voltage they will also in
phase. So, that gives us you that scoop about the proper connection for the CT, if we have that
polarity marking connect in accordance with that.

(Refer Slide Time: 20:39)

Now, what is the essential requirement from a good CT from the protection perspective? As we
know that these CT should give us the corresponding current in the secondary, which current it
will be a scaled-down current, but we know that all our analysis is based on the primary side that
is system side current.

In a numerical relay if the corresponding CT ratio is being multiplied to the secondary current then
the relay will produce the corresponding primary current that is what we have learned. Now, if that
is so we can say that all the calculations can be done in terms of the primary level or the system
level currents, for which we have only analysed in terms of that.

Now, this CT which we are discussing now is a magnetic coupling, it has a core and we know any
transformer core may there is a chance of saturation at times, if that happens to be there, then the
non linearity between B and H, Im and the induced EMF maybe lost beyond the saturation and
therefore you can say that the proposal may be, may not be there.

In that case, then there is challenge and so and that is what we say. Now, what is the expectation
of from a CT that always the corresponding secondary current should provide the proportionate
value of the primary current, the current transformer output signal which is the secondary current
input to the relay, it should be accurate reproduction of the corresponding primary current accurate
recovery production of the primary current.

Current transformer is for relaying are designed to have small errors during fault and while they
are performs during normal steady state currents may be compromised to certain accuracy. We
require that are correct reproduction for during particular is during all condition. Now, if we see
this plot the situation this is a pre-fault current, all inception point and the current goes very high
value primary current with a some decay and the secondary current is scaled down, this is
depending upon the ratio and then this is the secondary current for this case.

So, what happens you can say here if you see you plot the both you can say that in terms of scaling
and say that referring the corresponding primary current to the secondary current then only you
can plot otherwise it is a these very large value going beyond 5000, this is only within 100 A if
we convert the corresponding primary current to the secondary by dividing the primary current to
the suitable turns ratio and if you plot then you can observe that the secondary current is red one
and the primary current is blue one, then we say that this is exactly matching with this one.
So, it means that if this secondary current is fully available to the relay, relay will multiply simplify
the CT ratio and you can get the correct value of the primary side current. That is why the
reproduction is perfect. But this is ideal condition if the corresponding current becomes very large
and decaying DC will be there and so there is a possibility of saturation of the core this you can
say that reproduction of the primary may not be possible always, we will see that in the subsequent
lectures.

(Refer Slide Time: 24:42)

Now, I will first analyse the steady-state performance of the CT and then we will go to the transient
performance. So, at first we will draw the equivalent circuit diagram of this CT and from that you
can say that we will try to analyse how it performed during steady state. So, this is primary side
and this is the secondary side in the secondary side.

In the secondary side we have the burden and the burden impedance or resistance is already
mentioned is the impedance of all the relays, meters connected with the secondary plus the
corresponding lead you can see that impedance and that is the total burden to the CT. The CT has
certain leakage reactance for the secondary leakage reactance is Zx2 and primarily reactance is Z’x1.
Magnetizing impedance is Zm and then you can say that these are the windings where for every
turn secondary maybe number of turns, 1: n. So, with this you can say that the corresponding
current input to this primary which is the system conductor, transmission line conductor,
transformer conductor or so. In that the corresponding current flows, note one point here for the
CT the current which is being input to this primary is the conductor current of the system and that
does not depend upon the corresponding CT burden on so, that is independent of this corresponding
CT. So, that depends upon the load and the system operating condition. Therefore, the
corresponding current injected to the CT depends upon the system condition. The corresponding
input signal becomes a current source kind of thing. So, the equivalent becomes a current source.
Now, with this you can say that if we go with the transformation of this one referring everything
to the secondary because our measurements are on secondary, so we say consider the equivalent
becomes Zm, I1, Zx2, I2 and Zb, everything referred to the primary. We neglect here the Zx1 here
Z’x1 here and if you do not include here because as Z mentioned you can say that the current
considered is independent of the CT parameters and so. That depends on the system condition, so
that does not create any problem that you can say that have any influence on the I1 perspective,
and our objective here is that how the corresponding I2 can reproduces the corresponding I1, here
this I1 is already referred to the secondary side. Now, here you can see that this

𝐼1′
𝐼1 =
𝑛

𝑍𝑚 = 𝑛2 𝑍𝑚
′

So, in terms of this we call this equivalent of this primary and secondary refer to the secondary
side.

(Refer Slide Time: 27:58)

Now, with this we see you can say that the corresponding input current that I1 which is that of the
conductor current refer to the secondary, two components here one goes to the secondary side,
secondary you can say that winding perspective and one component in the magnetizing component
for the CT, any transformer has certain magnetic current component of current, because this is
having a magnetic core. So, this I2 goes to this burden, accordingly you can say that the burden
onto the CT will be there. So, for this case you can say that if we see the corresponding phasor
diagram, so if we say you can say that E b as a reference, so to the Eb in terms of accordance with
the Zb there we corresponding current of I2 and Zb as

𝐸𝑏 = 𝐼2 𝑍𝑏

Voltage across the magnetising branch is given by

𝐸𝑚 = 𝐸𝑏 + 𝐼2 𝑍𝑏

Im the magnetic current obtained from

𝐸𝑚
𝐼𝑚 =
𝑍𝑚

That is lagging almost 900 to Eb because of this Zm is very close having a large angle close to 900.
I1 is expressed as

𝐼1 = 𝐼2 + 𝐼𝑚

So, we got that currents and voltages from the parameters of the CT. Then we define error terms,
the that we can say as per unit current transformer error,

𝐼1 − 𝐼2 𝐼𝑚
𝜀= =
𝐼1 𝐼2

Here the I1 is refer to the secondary that means I1 corresponds to the primary current. Now, this I2
should be equals to I1. However, the I2 should be equals to I1, but that is not here and the reason
against here is nothing but this Im current. In an ideal case if this Zm becomes infinity then this Im
become 0, so I2 becomes equals to I1, so that is what desirable, now you can say it seems here that
if there will be Im the corresponding core then depend on the Im there will be inaccuracy you can
say that in the CT’s, it means that the relay which will be getting this current is I2, not the I1. So,
what is the error component here? The error is only solely responsible you can say that the
component is nothing but the magnetizing component of current. the error ɛ is small for larger Zb.
Theoretically ɛ is smaller for Zb =0 that is a sorted terminal agree, for very large value of Zm, Im
is smaller.

The ratio correction factor in divide I am say defined as R is defined as

1
𝑅=
1−𝜀

so what we say that whatever I2 at the relay if you multiply with R, the correction factor, then
you can get the corresponding correct value of I1. Why the inaccuracy? Because of this Im.
Therefore you require a correction factor to take into account this error, that is the epsilon and that
correction factor is your R, so if the corresponding I2 and by any means if it is known this
corresponding R value if we multiply that R value then you can say that the correct value can be
found out which will be proportional to the I1. Note, the ɛ you can see that the R also the complex
number all R we can say that in in terms of the phasor representation.

(Refer Slide Time: 33:40)

But many times the ɛ, R simply a magnitude is also adequate to for the expansion and so. Now, let
us take an example.

A current transformer with a turns ratio of 600: 5, turns ratio, 1: 200 you can say that within number
of turns, a secondary leakage impedance of (0.01+j0.15) Ω and a resistive burden of 1Ω. If
magnetizing impedance is (5+j17) Ω this part, find the correction factor R, compare the correction
factor for a inductive burden of j1Ω.

First you have considered for simple burden of resistive burden of 1 Ω and then we will go for the
inductive burden of 1 Ω. So, in this case in the equivalent circuit diagram what we have discussed
earlier that you know the different terminologies and note that in this case generally the Zx2 is
much smaller.

Zm = 5.0+j17 Ω, Zx2 =0.01+j0.15 Ω , Zb =1.0 Ω

𝑍 ×(𝑍𝑥2 +𝑍𝑏 ) (5.0+𝑗17)(0.01+𝑗0.15+1.0)

𝐸𝑚 = 𝐼1 (𝑍𝑚 = 𝐼1 (5.0+𝑗17+0.01+𝑗0.15+1.0) = 𝐼1 (0.996∠11.370 )
𝑚 +𝑍𝑥2 +𝑍𝑏 )

𝐸𝑚 𝐼1 × 0.996∠11.370
𝐼𝑚 = = = 𝐼1 × 0.056∠ − 62.240
𝑍𝑚 5.0 + 𝑗17

𝐼𝑚
Per unit CT error 𝜀 = = 0.056∠ − 62.240
𝐼1

1 1
Correction factor 𝑅 = 1−𝜀 = 1−0.056∠−62.240 = 1.025∠ − 2.910

(Refer Slide Time: 36:12)

Now, for the inductive load For Zb = j1.0 Ω

 𝑍 ×(𝑍𝑥2 +𝑍𝑏 ) (5.0+𝑗17)(0.01+𝑗0.15+𝑗1.0)
𝐸𝑚 = 𝐼1 (𝑍𝑚 = 𝐼1 (5.0+𝑗17+0.01+𝑗0.15+𝑗1.0) = 𝐼1 (1.082∠88.340)
𝑚 +𝑍𝑥2 +𝑍𝑏 )

𝐸 𝐼1 ×1.082∠88.34𝑜
and 𝐼𝑚 = 𝑍𝑚 = = 𝐼1 (0.061∠14.930 )
𝑚 5.0+𝑗17

𝐼𝑚
𝜀= = 0.061∠14.930
𝐼1

1 1
𝑅= = = 1.062∠0.9570
1−𝜀 1−0.061∠14.930

So, if you compare both the case you can say that the ɛ is higher for the inductive case and so you
can say that is also the corresponding correction factor become higher. In general for the resistive
burden we will have lesser or compared to same amount of inductive burden.

So, if you know the corresponding inductive burden you can say that what are the corresponding
ɛ error part, then if you are you can say that system will be connected to more resistive of the same
value of the inductance, then you say that the corresponding error will be less than that. So, always
you can say that the corresponding error to be as small as possible.

(Refer Slide Time: 37:26)

Note, class of CT as per the IEEE standard, so CT classification we can see we are discussing on
protection perspective, but CT is available for both protection class and metering or measurement
class, so we will discuss only on the protection class perspective. The IEEE class designations of
a CT it defines in terms of two integer parameters C and T and for example 10C400 or 10T300.
So, we have integer before the C and 400 after it and we have integer before the T and after consider
T also. Now, the letter C in the class imply the transport design is such that the CT performance
can be calculated, C for calculated and the T signifies some uncertainties in the transformer design
and the performance of the CT must be determined by the testing the CT’s. That is T for the testing
and C for the calculation. So, we can calculate the corresponding performance of the CT based on
the data sheet provided by the manufacturer. Now, what these 10 and 400 and 300 in this tool
reveal? The first integer 10 here described the upper limit on the error made by the CT when the
voltage at its secondary terminal is equal to the second integer, when the voltage in the secondary
terminals is equal to second integer that this voltage will be 400 V while the current in the
transformer is 20 times the rated value, when this is being evaluated the corresponding transformer
you can say that current will be 20 times of the rated value. So, if it is a 5 A CT for the
corresponding current should be 100 A in this secondary that is you can say and what is this 10?
The 10 reveals about the error you can say made by this in terms of 10 % error, percentage of this
error and this 400 reveals the corresponding voltage in the burden error when you can say the
corresponding current is 20 times of the rated one. So, if we see these 10C400 here this corresponds
to percent error and this current correspond to the secondary voltage 20 time the corresponding
current becomes 20 times of the it's rated value.

Accuracy is specified by percentage error and you can say that should not be considered a accuracy
you can say that for that particular class as per the IEEE standard and the ratio you can say that
error is always you can say that 10 in general, but there are other standards also IEC standard in
all these things, so they have their own you can say that definitions and all this things.

But there are tables also available in terms of the competition that actually standard this class or
this rating what is the corresponding IEC you can say that rating and so. The IEC standard burden
for the relays are 1, 2, 4, 8, Ω standard burdens all with an impedance angle of 600. However, the
CT is classified by a voltage across the burden that is we say this voltage, so the corresponding
voltage becomes in terms of 100 V, 200 V, 400 V and 800 V this part you can say that which are
talking about in terms of this sequence of impedances corresponds to 100 V, 200 V, 400 V and
800 V this is what as per the IEEE standard.
(Refer Slide Time: 41:07)

Now, you go for a burden calculation example,

Example: A current transformer of 500:5 with a primary current of 6000 A of CT class of 10C400,
what should be the value of Zb

Solution: CT secondary winding is rated at 5 A secondary, this corresponds to a maximum

secondary current of

20×5= 100 A.

The 10C400 CT will have an error of ≤10% at a secondary current of 100 A for burden
impedances which produce 400 V or less at its secondary terminals.

The magnitude of the magnetizing impedance for maximum error

400
= 40 Ω .
𝐼𝑚 (=10% 𝑜𝑓 100𝐴)

For a primary current of 6000 A, the nominal secondary current will be

6000× (5/500) = 60 A.

With a maximum error of 10 %, this will allow a magnetizing current of 6 A. At this magnetizing
current, it will have a maximum secondary voltage

=40x6=240 V.
The primary current is 60 A, the maximum burden impedance which will produce 240 V at the
secondary is

240
= = 4.44 Ω.
(60−6)

And that is the maximum one if the corresponding burden between is less than this burden, then
you can say that the 10 % in the error will be less than 10 %, if it is more than that the error will
be more than 10 %. So, that will reveal that in the for that kind of CT that how much you can say
that burden impedance will be there. So, the corresponding lead and the corresponding number of
relay and so the way it should be connected such that you can say that the burden impedance does
not accept this 4.44 Ω.

(Refer Slide Time: 44:49)

Another example consider a CT with a turns ratio of 500:5 and the magnetizing characteristic for
different ratios IS given below. It is required to calculate the current in its secondary winding for
a primary current of 4000 A, if the total burden impedance is (8+j3) Ω and the secondary leakage
impedance is negligible. The impedance angle of the magnetizing branch is 65 0.
Solution: Current source of 4000 × 5/500 = 40.0 A (secondary side) in parallel with the burden,
and connected across the nonlinear impedance Z𝑚 .

The equivalent Thevenin voltage source of 40.0 × (8+j3) = 341.76 ∠20.560 V, in series with the
burden. Since the impedance angle of Z𝑚 is known to be 650, the magnetizing current I𝑚 and the
secondary voltage E2 can be expressed in terms of the magnitude of Z𝑚 with the Thevenin voltage
as the reference phasor:

341.76
Im =
[(|𝑍𝑚 | × (0.423 + j0.906)) + (8 + j3)]

E2 = Im Z𝑚

(Refer Slide Time: 45:53)

These two equations may be solved to produce values of E2 and Im in terms of |Z𝑚 | as the
parameter (Table).

Z𝑚 (ohm) Im (A) E2 (V)

∞ 0 341.76

100 3.22 321.60

10 19.90 198.98

(Refer Slide Time: 47:15)

(Refer Slide Time: 48:03)

So, why did you do that you plot you can say that the corresponding E 2 verses the corresponding
Im line and that line you can say that crosses the considered at this point you can say that at this
point to this the corresponding magnetizing characteristics and we got the value we can say that
for that situation for that burden the corresponding 19 you can say that 19 A and then the
corresponding value of we can say that the secondary voltage to be 250 V

Plotting the curve of these values on the non-linear characteristics, it is found to intersect the
magnetizing characteristic at Im = 19 A, E2 = 250 V.

I1 = 40.0∠00 (in that case, Eth = 341.76∠20.560), Im = 19.0∠ − 30.450 and I2 =

25.50∠22.170. The error ∈ =0.475∠ −30.450 and the ratio correction factor R = 1.568∠ −
22.180 . CT is in severe saturation at this current. Burden must be used with much smaller value
reveals that the CT’s in a severe saturation current because the ɛ value is very large here you see
here as compared to the earlier discussion we made and that can said why the situation, that because
it goes to the saturation in a region in the BH curve and there is the region of the corresponding
error becomes this, what the error reveals that the corresponding I2 which is being seen by the
corresponding relay is having a much you can say that smaller value because of this significant Im
it is compared the injected current I1. So, the burden should be reduced is it seems to be very high.
(Refer Slide Time: 49:40)

So, in overall our remarks you can say that is that a CT with a standard burden having a impedance
angle of 600 produces less than 10% error at 20 times current and it resistive burden of the same
will produce less than 5% error, this is a general colour markings you can say that the calculation
is we are that resistive burden will having less error as compared to the similar same value of
inductive burden. CT selection in terms of IEEE standard suggests their CT’s for relaying be
applied such that the maximum symmetrical fault current does not exceed 20 times the CT current
rating and that the burden voltage does not exceed the accuracy class voltage of the CT. So,
preference will be that the CT current does not exceed you can see that 20 times of the CT rating
and the burden you can say it should be solved that you can say that the corresponding class being
satisfied. When CT saturated the magnetizing current increases a substantial resulting more error
that we see for the burden should be incompatible to the requirement for the accuracy class of the
CT. So, this is all on CT when say that steady-state behaviour. Thank you.