Answer.1. Dangerously high voltage develop across secondary
Explanation:
The phase angle error in a C.T. is given by the relation
\(\frac{180}{π} [\frac{I_m}{nI_s}]\)
where
θ is the phase angle error
Im is the magnetising component of the excitation current
Is is the secondary winding current
The phase-angle error is largely dependent upon the value of the magnetizing component Im of the exciting current.
52. A 50Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a pure resistance burden of 1 Ω. The magnetizing ampere-turns is 200. The phase angle between the primary and reversed secondary current is
4.6°
85.4°
94.6°
175.4°
Answer.1. 4.6°
Explanation:
N1 = 1, N2 = 500, Is = 5A, R2 = 1Ω
Magnetizing ampere turns = 200 AT
Io = Magnetizing ampere turns/Primary turns = 200 A
53. For eliminating errors in power measurement in the current transformer, there must be a phase difference of _______
90°
45°
180°
270°
Answer.3. 180°
Explanation:
For a perfect (ideal) current transformer, the phase difference between the primary and reversed secondary phasors is zero. But for an actual transformer, there is always a difference in phase between the two phasors due to the fact that the primary current has to supply the components of the exciting current. While power measurements, it is required that the secondary current of CT is displaced exactly by 180° from the primary current.
For eliminating errors in power measurement, there must be a phase difference of 180° between the primary and the secondary currents.
54. A CT is connected in ______ with the line
Series
across
Both A and B
Not connected
Answer.1. Series
Explanation:
The primary CT is connected in series with a power line. So current through its primary is nothing but the current flows through that power line. The primary current of the CT does not depend upon whether the load or burden is connected to the secondary & also not on the impedance value of the burden.
55. Errors in a C.T. can be minimised by _________
Making use of laminations
Having low reactance
Increasing the secondary winding turns
Decreasing the primary winding turns
Answer.1. Series
Explanation:
In order to minimize the exciting ampere-turns required to reduce these errors, the core must have a small iron loss and a low reluctance path. The flux density should be as low as possible.
The length of the magnetic path should be as small as is consistent with good mechanical construction and with insulation requirements. This results in a low reluctance path. Errors in a C.T. can be minimized by having low reactance (high mu) and small iron loss.
56. A current transformer (CT) of turns ratio 1 : 248 is rated as 1000 / 4 A, 20 VA. The core loss and magnetizing components of the primary current are 4 A and 8 A under rated conditions, the phase angle error for the rated burden and the rated secondary current at 0.8 power factor lagging is
57. A bar primary current transformer of rating 1000/1 A, 5VA, UPF has 995 secondary turns. It exhibits zero ratio error and phase error of 30 minutes at 1000 A with the rated burden. The watt loss component of the primary excitation current in ampere is ______
10 A
5 A
2 A
15 A
Answer.2. 5 A
Explanation:
Given
Ns = 995
Ratio error = 0
phase error = 30 minutes
I = 1000 A
n = 1000
as ratio error is zero, we can write R = kn
n = Turns ratio secondory to primary
q = 30 min = 0.5 degree
UPF = resistive burden
(δ = 0)
θ = (180 Im)/(π × n × Is)
Where, Im = magnetising current, Is = secondary current of CT.
⇒ 0.5 = (180 Im)/(π × 1000 × 1 )
⇒ Im = 8.68 A
We know that;
R = Ns + (Ic/Is)
⇒ 1000 = 995 + (ic/1)
⇒ Ic = 5 A
58. A current transformer (CT) has a rating of 100/5 A. Its magnetizing and loss components of the exciting current are 6 A and 4 A respectively and its burden is purely resistive. The transformation ratio of CT at rated current is:
20
24
26
20.8
Answer.4. 20.8
Explanation:
Transformation Ratio of CT
n = Ip/Is
Given,
n = 100/5 = 20
Is = 5∠0°
Is = 5 A
Ip = nIs + Io
where
Is = Secondary current
Ip = Primary current
I0 = Excitation current.
Im = Magnetizing component of I0.
Iw = Core loss component of I0.
Ip = nIs + (Iw + jIm)
Ip = 20 × 5 + (4 + j6)
Ip = 104 + j6
Ip = 104.172
So, the Transformation Ratio of CT
n = 104.72/5 = 20.83
59. A transformer is an electrical device that transfers electrical energy from one electric circuit to another, without changing:
Capacitance
Current
Frequency
Voltage
Answer.3. Frequency
Explanation:
A transformer is a static device that converts electrical power from one circuit to another without changing its frequency. So that input and out supply frequency always remain the same.
Thus for a given 50 Hz input supply frequency, the output supply frequency is 50 Hz.
60. A current transformer (CT) of turns ratio 1 : 248 is rated as 1000 / 4 A, 20 VA. The core loss and magnetizing components of the primary current are 4 A and 8 A under rated conditions. The ratio error for the rated burden and the rated secondary current at 0.8 power factor lagging is
