Previous Year Questions- Instrument Transformers | Electrical and Electronic Measurements - Electrical Engineering (EE) PDF Download

Q1: The table lists two instrument transformers and their features:
The correct matching of the two column is (2024)
(a) X matches with P and Q; Y matches with R and S.
(b) $X$X matches with P and R; Y matches with Q and S.
(c) $X$X matches with Q and R; Y matches with P and S.
(d) X matches with Q and S; Y matches with P and R.
Ans: (c)

Q2: A 200/1 Current transformer (CT) is wound with 200 turns on the secondary on a toroidal core. When it carries a current of 160 A on the primary, the ratio and phase errors of the CT are found to be -0.5% and 30 minutes respectively. If the number of secondary turns is reduced by 1 new ratio-error(%) and phaseerror( min) will be respectively (2006)
(a) 0.0, 30
(b) -0.5, 35
(c) -1.0, 30
(d) -1.0, 25
Ans: (a)
Sol: Nominal ratio k_n = 200/1 = 200
Primary current = I_P = 160A
Actual ratio
%Ratio error Actual ratio = R₁ = 201
n₁ = turn ratio = 200
when number of secondary is reduced by 1

N_s = 199
N₂ = 199
Dividing equation (ii) and (iii),
%Ratio error = Phase angle error
Reduction of one or two turns of the secondary winding, no doubt, reduces the ration error, but it has no effect on the phase angle error.

Q3: A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1Ω. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is 175.4°. The core flux in the CT under the given operating conditions is (2004)
(a) 0
(b) 45 μWb
(c) 22.5 m Wb
(d) 100.0mWb
Ans: (b)
Sol: Z_s = Burden = 1Ω
Voltage induced in the secondary
E_s = I_S × Z_S = 5 × 1 = 5V
Es = 4.44fϕ_mN_s

Q4: A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1Ω. The magnetizing ampere-turns is 200. The phase angle between the primary and second current is (2004)
(a) 4.6°
(b) 85.4°
(c) 94.6°
(d) 175.4°
Ans: (a)
Sol: Phase angle between primary and secondary current
No. of turns in primary = N_p = 1
No. of turns in secondary = N_s = 500
n = turn ration=turn ratio = N_s/N_p = 500
I_s = secondary winding current = 5A
Magnetizing ampere turn = 200

Q5: A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure resistance of 1 Ω and it draws a current of 5 A. If the magnetic core requires 250 AT for magnetization, the percentage ratio error is (2003)
(a) 10.56
(b) -10.56
(c) 11.8
(d) -11.8
Ans: (b)
Sol: Nominal ratio = 500/5 = 100
Number of turn in primary N = 1 (primary bar)
Magnetizing current,
Secondary current,
I_s = 5An = 500/5 = 100
Primary current,
⇒ I_P = 559.017A  
Actual ratio
Ratio error