Electrical and Electronic Measurements& Instrumentation
10EE35
Testing of Current Transformer
There are three methods:
1. Mutual Inductance method: This is an absolute method using null deflection.
2. Silsbee’s Method: This is a comparison method. There are two types; deflectional and
null.
3. Arnold’s Method: This is a comparison method involving null techniques.
Silsbee’s Method: The arrangement for Silsbee’s deflectional method is shown in Fig.1.
Here the ratio and phase angle of the test transformer ‘X’ are determined in terms of that
of a standard transformer ‘S’ having the same nominal ratio
Procedure:
The two transformers are connected with their primaries in series. An adjustable
burden is put in the secondary circuit of the transformer under test.
An ammeter is included in the secondary circuit of the standard transformer so
that the current may be set to desired value. W1 is a wattmeter whose current coil is
connected to carry the secondary current of the standard transformer. The current coil of
wattmeter W2 carries a current I which is the difference between the secondary currents
of the standard and test transformer. The voltage circuits of wattmeters are supplied in
parallel from a phase shifting transformer at a constant voltage V.
The phasor diagram is shown in Fig. 2
Electrical and Electronic Measurements& Instrumentation
10EE35
Fig.2. Phasor diagram of Silsbee’s method
1. The phase of the voltage is so adjusted that wattmeter W1 reads zero. Under these
conditions voltage V is in quadrature with current Iss . The position of voltage phasor for
this case is shown as Vq .
Reading of wattmeter, W1 W1q = Vq Iss cos 90° = 0.
Reading of wattmeter, W2
W2q = Vq × component of current I in phase with Vq = Vq Iq
= Vq Isx sin (θx – θs)
Where θx = phase angle of C.T. under
test, θs = phase angle of standard C.T.
= W1p– VIsx cos (θx – θs) ≈ W1p– VIsx
As (θx – θs) is very small and, therefore, cos (θx – θs) =
1 For above, VIsx = W1p - W2p.
Actual ratio of transformer under test Rx = Ip
/Isx. Actual ratio of standard transformer Rs = Ip
/Iss.
. The phase of voltage V is shifted through 90° so that it
occupies a position VP and is in phase with Iss .
Reading of wattmeter W1, W1p = Vp Iss cos θ = Vp Iss.
Electrical and Electronic Measurements&
Instrumentation
10EE35
Reading of wattmeter W2, W2p = Vp ×component of current
Ip = Vp [Iss – Isx cos (θx – θs)]
If the voltage is kept same for both sets of
readings, then V= Vp = Vq.
We have W2q = VIsx sin (θx – θs), W1p = Viss
W2p = V[Iss – Isx cos (θx – θs)]= VIss – VIsx cos (θx – θs)]
W
2
R 1 p
R
x
s
W
1p
W
W
2
q
)
) VI ss − W2 p
−θ
x s
sin(θ
1 p − W2 p
VI sx
cos(θ − θ
x s
VI sx
W
W
2q
∴
tan(θ
I in phase with Vp = Vp ×
−θ
x s
or
(θ x − θs
W
W −W
1
2
p
p
VI sx
2q
1 p − W2
p
rad
Or phase angle of test transformer,
W
W
θ
2q
2q
θ rad ≈
θ rad
s
s
W
1p
−W
2p
W
1p
as W2p is very small. Hence if the ratio and phase angle errors of standard transformer are
known, we can compute the errors of the test transformer. W2 must be a sensitive
instrument. Its current coil may be designed for small values. It is normally designed to
carry about 0.25A for testing CTs having a secondary current of 5A.
Electrical and Electronic Measurements& Instrumentation
10EE35
Problem No.1
Two current transformers of the same nominal ratio 500/5 A, are tested by Silsbee’s
method. With the current in the secondary of the transformer adjusted at its rated value,
the content in the middle conductor I = 0.05e-j126.9° A expressed with respect to current
in the secondary of standard transformer as the reference. It is known that standard
transformer has a ratio correction factor (RCF) of 1.0015 and phase error +8’. Find RCF
and phase angle error of transformer under test.
Solution:
Nominal ratio = 500/5 =100, Iss= 5 A
Since Iss is the reference, Iss= 5+j0
I = 0.05e-j126.9°
= 0.05(cos 126.9° -j sin 126.9°)
= -0.03 – j 0.04
Now, current in the secondary of test transformer
Isx=Iss- I = 5 +j 0 – (0.03 – j 0.04) = 5.03 + j 0.04
or Isx ≈ 5.03 A
Angle between Isx and Iss= 0.04/5.03 rad = 27.3’.
Phase angle between Isx reversed and Ip=+27.3’+8’ = +35.3’.
Ratio correction factor RCF of standard transformer = 1.0015
Therefore, actual ratio of standard transformer
Rs= RCF × nominal ratio = 1.0015 × 100 = 100.15
Primary current Ip= Rs ×Iss= 100.15 × 5 = 500.75A
Actual ratio of transformer under test
Rx= Ip/Is = 500.75/5.03 = 99.55
Ratio correction factor of test transformer RCF =
Actual ratio/nominal ratio = 99.55/100 = 0.9955.