SYSTEM TRANSDUCERS

CT Nameplate

core 1:Busbar(discriminating zone)

core 2:Busbar(check zone)
core 3:Main Prot(CD)
core 4:Disctance+metering
CT
VT Nameplate
The most common type of C.T. construction is the “DOUGHNUT” type.
It is constructed of an iron toroid, which forms the core of the
transformer, and is wound with secondary turns.
VA = VI. V= IR, So 0.2 = I*I*R.
0.2/25 = 0.008 ohms
diameter
According to IEC, Knee Point Voltage of a
Current Transformer is defined as the
voltage at which a 10 % increase in voltage
of CT secondary results in a 50 % increase
in secondary current.
 100:5 C.T. Secondary Winding Resistance (DCR) = 0.062 ohm
Resistance of Cable from C.T. to Relay and back = 0.1 ohms
Resistance of Relay Coil = 0.02 ohms
Total Resistance = 0.182 ohms

0.062 0.02

0.1(loop)

If we have a fault of 2,000 amps and the C.T. ratio is 100:5 then the C.T.
secondary current is 100 amps. Therefore we will produce a voltage of 100 amps x
0.182 ohms = 18.2 Volts. To prevent CT saturation, select a CT with a knee point
above 18.2 Volts.
The name plate is for 3-Core CT
with middle taps. For this CT the
knee point voltage for PS-Class core
is Vk >= 120V for 3S1-3S3 winding
and Vk >=60V for 3S1-3S2 winding.
Also the excitation current should
less than or equal to 100mA at VK/4
for anyone of Core-3 windings
780-102 is a 1000 to 5 CT, Class C100
1000:5 C.T. Secondary Winding Resistance (DCR) = 0.32 ohm
Resistance of Cable from C.T. to Relay and back = 0.1 ohms
Resistance of Relay Coil = 0.008 ohms
Total Resistance = 0.428 ohms

0.32
0.008
0.1 (loop)

If we have a fault of 20,000 amps and the C.T. ratio is 1000:5 then the C.T.
secondary current is 100 amps. Therefore we will produce a voltage of 100
amps x .428 ohms = 42.8 Volts. To prevent CT saturation, select a CT with a
knee point above 42.8 Volts.
What happens if the fault current is 40,000 amps?
No matter how high the voltage is,
the flux produced is limited due to core
saturation.
Let us take flux as reference. EMF Es and Ep lags behind the flux by 90°. The
magnitude of the phasors Es and Ep are proportional to secondary and primary
turns. The excitation current Ie which is made up of two components Im and Ic
From above phasor diagram it is clear that primary current Ip is not exactly
equal to the secondary current multiplied by turns ratio, i.e. KnIs.

This difference is due to the primary current is contributed by the core

excitation current. The error in current transformer introduced due to this
difference is called current error of CT or some times ratio error in current
transformer.
EXAMPLE

In a 300/5A CT, the measured secondary current of a primary

current of 300 is 4.9 A. Calculate the CT ratio error.

Solution

300
Kn = = 60
5

Referred to secondary: Referred to primary:

Ip Kn Is − I p
Is − Current ratio error = ×100%
Kn Ip
Current ratio error = × 100%
Ip 60(4.9) − 300
Kn = ×100% = −2%
300
4.9 − 300
= 60 × 100% = −2%
300
60
Model 203 CTs mounted in rear bus compartment
 60 minute ' = 1 degree °

Typical CT
Models
112
113
114
115
117

Polyester Taped Bushing CT on

Outdoor Circuit Breaker
Evaluation of Ct error
 Zeq’

23.73 V

7.06 V

2.91 V

0.25A 0.4A 20A

 EXAMPLE
An over current relay set to operate at 8A is connected to a 100/5A CT with Is = 8A. Will the
relay detect a 200A primary fault current if the burden ZB
(a) 0.8 Ω
(b) 3.0 Ω
Use excitation current curve and secondary resistance table for multi-ratio CT.
SOLUTION
(a) IS = 8A (overcurrent relay setting)
IP = 200A (fault current), 100/5A CT  Zeq’ = 0.082
Is = 8A
Z total = Z eq '+ Z b = 0.082 + 0.8 = 0.882Ω
Ip’ Is Zeq’
Es = I s × Z total = 8 × 0.882 = 7.06 V
Io’
from the curve, I o ' = 0.4 A Es
Ze’ ZB

I p ' = I s + I o ' = 8 + 0.4 = 8.4 A

100
I p = I p '× K n = 8.4 × = 168 A
5

IP = 168A produces IS = 8A  relay will operate

So, if IP = 200A  relay will operate
 EXAMPLE
Consider a 13.2kV, 3-phase feeder which is supplying a 3-phase load of 10MVA at 1.0 power factor (pf) shown
in Figure Q2-1. Each phase is associated with a 500/5 current transformer (CT) feeding a 5A ammeter which
has a total burden of 10VA at 1.0 pf.

a) Determine the ammeter reading and voltage across the ammeter if the excitation current of the CT, Io’ is
0.02A.
(5 marks)
b) Calculate the voltage that would occur in the secondary circuit of the CT if the ammeter is accidentally
opened. (Given: Excitation branch of the CT, Rc=150Ω and Xm=50Ω).
Load (2 marks)
Ip’ Is Zeq’
5A
VL − L = 132kV CT A Ammeter Io’
Ze’ ZB
Es A

Figure Q2-1
13.2𝑘𝑘𝑘𝑘
𝑉𝑉ϕ = = 7.621kV
3
Ammeter reading = 𝐼𝐼𝑆𝑆 = 4.354A
𝑆𝑆3ϕ = 3𝑉𝑉ϕ 𝐼𝐼ϕ * @ 𝑆𝑆 = 𝑉𝑉ϕ 𝐼𝐼ϕ *
Voltage Across ammeter = 𝐼𝐼𝑆𝑆 × 𝑍𝑍𝐵𝐵
𝐼𝐼ϕ =
10𝑀𝑀𝑀𝑀𝑀𝑀�

7.621𝑘𝑘𝑘𝑘
3
= 437.39A=𝐼𝐼𝑃𝑃 P = S x cosθ
P = 10 x 1
𝐼𝐼ϕ 437.39A
P = 10
𝐼𝐼𝑃𝑃′ = = = 4.374𝐴𝐴
𝐾𝐾𝑛𝑛 500�
5
10 = 𝐼𝐼2 𝑍𝑍𝐵𝐵
Given 𝐼𝐼𝑂𝑂′ = 0.02𝐴𝐴, 10
𝐼𝐼𝑝𝑝′ = 𝐼𝐼𝑂𝑂′ + 𝐼𝐼𝑆𝑆  𝐼𝐼𝑆𝑆 = 𝐼𝐼𝑃𝑃′ − 𝐼𝐼𝑂𝑂′ 𝑍𝑍𝐵𝐵 = 2 = 0.4
5
= 4.354A
CT error =
𝐼𝐼𝑆𝑆 −𝐼𝐼𝑝𝑝′
× 100%
Therefore, 𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐼𝐼𝑆𝑆 × 𝑍𝑍𝐵𝐵
𝐼𝐼𝑝𝑝′

=
4.354−4.374
4.374
× 100% = 4.354 × 0.4
= - 0.46%
= 𝟏𝟏. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
(b) Ip’ Is Zeq’

Io’
Ze’ ZB
Es

Given Rc = 150Ω and Xm = 50Ω

150×𝑗𝑗𝑗𝑗
Z = 150//j50 = = 47.43<71.56𝑂𝑂
150+𝑗𝑗𝑗𝑗

Es = 𝐼𝐼𝑂𝑂 ′ × 𝑍𝑍 = 0.02×(47.43<71.56𝑂𝑂 )
= 0.9486<71.56𝑂𝑂 V @ 0.3+j0.8998V

Remember: The excitation current Ie which is made up of two components Im and Ic.
If both C1 and C2 values are taken
VT Nameplate
 EXAMPLE 2
The equivalent circuit of a CVT is shown in figure 2. The values of C1
and C2 are 0.0018𝜇𝜇𝜇𝜇 and 0.0186𝜇𝜇𝜇𝜇 respectively. Tuning inductor has
an inductance of 497H and resistance of 4620Ω. Xm of the VT referred
to 6.6 kV side is 1MΩ , core loss = 20 watts per phase, VA burden =
150VA per phase. Value of Cm for compensating the current drawn by
Xm is equal to 3.183 x 10-9F.
a) Verify the appropriateness of
choice of L and Cm .

b) Find out the nominal value of V/V2

c) If the frequency drops from 50Hz

to 47Hz, what would be the values
of ratio error and phase angle
error?
Figure 2
 Answer (a)
If C1 = 0.0018𝜇𝜇𝜇𝜇 and C2 = 0.0186𝜇𝜇𝜇𝜇 and then the value of tuning
inductor, L is given by
1
𝐿𝐿 =
ω2 (𝐶𝐶1 + 𝐶𝐶2 )
where ω = 2𝜋𝜋𝜋𝜋 and 𝑓𝑓 = nominal frequency. Thus,
1
𝐿𝐿 = = 496.7𝐻𝐻
2𝜋𝜋 × 50 2 (0.0018𝜇𝜇𝜇𝜇 + 0.0186𝜇𝜇𝜇𝜇)

496.7H which is equal to the given value of L. Now, Xm = 1MΩ

and Cm has to be in parallel resonance with Xm . Therefore,
1 1 1 −9
𝑋𝑋𝑚𝑚 = , 𝐶𝐶𝑚𝑚 = = 6
= 3.183 × 10 𝐹𝐹
ω𝐶𝐶𝑚𝑚 ω𝑋𝑋𝑚𝑚 (2𝜋𝜋 × 50) × 1 × 10

The value is also same as the selected value of Cm. Hence, the
selection of both L and Cm is appropriate.
 Answer (b)

Find out the nominal value of V/V2

𝑉𝑉 𝐶𝐶1 + 𝐶𝐶2 0.0018𝜇𝜇𝜇𝜇 + 0.0186𝜇𝜇𝜇𝜇

= = = 11.33
𝑉𝑉2 𝐶𝐶1 0.0018𝜇𝜇𝜇𝜇

132
𝑉𝑉 = 11.33𝑉𝑉2 = 11.33 × 6.6𝑘𝑘𝑘𝑘 = 74.778𝑘𝑘𝑘𝑘 ≈ 𝑘𝑘𝑘𝑘
√3
Thus, this VT is connected to a 132 kV bus.
 Answer (c)
𝑉𝑉22 𝑉𝑉22 (6600)2
Core loss = 20W = , Therefore 𝑅𝑅𝑚𝑚 = = = 2.178 × 106 Ω
R𝑚𝑚 20𝑊𝑊 20𝑊𝑊
𝑉𝑉22 (6600)2
VA burden = 150VA (resistive), 𝑅𝑅𝑏𝑏 = = = 2.904 × 105 Ω
150𝑉𝑉𝑉𝑉 150𝑉𝑉𝑉𝑉
𝑋𝑋𝑚𝑚 = 1 × 106 Ω 𝑎𝑎𝑎𝑎 𝑓𝑓 = 50ℎ𝑧𝑧, 2𝜋𝜋𝜋𝜋𝐿𝐿𝑚𝑚 = 1 × 106 Ω 𝐿𝐿𝑚𝑚 = 3183.1𝐻𝐻
The equivalent circuit can be represented as shown Figure 3 below.

Figure 3
 Answer (c)
The frequency of interest is 47Hz. Hence values of Xm and other
impedance can be calculated at 47Hz. Figure 3 can be simplified
as Figure 4

Figure 4

Where 1 1 𝑗𝑗 1
= − + 𝑗𝑗ω𝐶𝐶𝑚𝑚 +
𝑍𝑍 𝑅𝑅𝑚𝑚 𝑋𝑋𝑚𝑚 𝑅𝑅𝑏𝑏
1 1 𝑗𝑗 1
= − + 𝑗𝑗𝑗𝜋𝜋𝜋𝜋𝐶𝐶𝑚𝑚 +
𝑍𝑍 𝑅𝑅𝑚𝑚 2𝜋𝜋𝜋𝜋𝐿𝐿𝑚𝑚 𝑅𝑅𝑏𝑏
 Answer (c)
1 1 𝑗𝑗 −9
1
= − + 𝑗𝑗𝑗𝜋𝜋(47)(3.183 × 10 ) +
𝑍𝑍 2.18 × 106 2𝜋𝜋 47 3183.1 2.904 × 105
1
= 0.459 × 10−6 − 𝑗𝑗𝑗.064 × 10−6 + 𝑗𝑗𝑗.94 × 10−6 + 3.44 × 10−6
𝑍𝑍
1
= 3.902 − 𝑗𝑗𝑗.124 × 10−6 = 3.904 × 10−6 ∟ −1.820
𝑍𝑍
1 0 Ω = 256,018.32 + 𝑗𝑗𝑗,135.15Ω
𝑍𝑍 = = 256,147.5∟1.82
3.904 × 10−6 ∟ −1.820
𝑉𝑉𝑡𝑡𝑡
𝐼𝐼𝑡𝑡𝑡 =
𝑗𝑗
𝑅𝑅 + 𝑗𝑗ω𝐿𝐿 − + 𝑍𝑍
ω𝐶𝐶𝑒𝑒𝑒𝑒
6600∟00
𝐼𝐼𝑡𝑡𝑡 =
𝑗𝑗
4620 + 𝑗𝑗𝑗𝜋𝜋 × 47 × 496.7 −
2𝜋𝜋 × 47 × 0.0204𝜇𝜇𝜇𝜇 + 256,018.32 + 𝑗𝑗𝑗,135.15Ω
 Answer (c)
6600∟00
𝐼𝐼𝑡𝑡𝑡 =
4620 + 𝑗𝑗146,768.9 − 𝑗𝑗𝑗𝑗𝑗,994 + 256,018.32 + 𝑗𝑗𝑗,135.15Ω

6600∟00
𝐼𝐼𝑡𝑡𝑡 =
260,638.32 − 𝑗𝑗𝑗𝑗,089.84

6600∟00
𝐼𝐼𝑡𝑡𝑡 = A
260,874.14∟−2.440

𝑉𝑉𝑇𝑇 = 𝐼𝐼𝑡𝑡𝑡 × 𝑍𝑍
6600∟00 0Ω
𝑉𝑉𝑇𝑇 = × 256,147.5∟1.82
260,874.14∟ −2.440
𝑉𝑉𝑇𝑇 = 6,480.42∟4.260
 Answer (c)
𝑉𝑉𝑡𝑡𝑡 − 𝑉𝑉𝑇𝑇
Amplitude or ratio error of a CCVT is equal to 𝑉𝑉 × 100 where V is the Thevenin
𝑡𝑡𝑡 th
equivalent voltage source is nothing but open circuit emf. VT is the terminal voltage on
load. Hence % ratio error

6600−6480.42
%𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒, α = = 1.81%
6600

pℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒, β = 4.260

Clearly, the phase angle error is on the higher side.