# 250+ TOP MCQs on Advanced Problems on C.T. and P.T. and Answers

Tricky Electrical Measurements Questions and Answers on “Advanced Problems on C.T. and P.T.”.

1. A 50 Hz, bar primary CT has a secondary with 800 turns. The secondary supplies 7 A current into a purely resistive burden of 2 Ω. The magnetizing ampere-turns are 300. The phase angle is?
a) 3.1°
b) 85.4°
c) 94.6°
d) 175.4°
Clarification: Secondary burden is purely resistive and the resistance of burden is equal to the resistance of the secondary winding; the resistance of secondary winding = 1Ω. The voltage induced in secondary × resistance of secondary winding = 7 × 2 = 14V. Secondary power factor is unity as the load is purely resistive. The loss component of no-load current is to be neglected i.e. Ie = 0. IM = 300 A.
Secondary winding current IS = 7 A
Reflected secondary winding current = n IS = 5600 A
∴ tan θ = (frac{I_M}{nI_S}). So, θ = 3.1°.

2. A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 Ω. The magnetizing ampere-turns are 200. What is the ratio error at full load, if the iron loss is 1.5 W and magnetizing mmf is 100 A?
a) Zero
b) 45 μWb
c) 25.5 μWb
d) 100 μWb
Clarification: Voltage induced in the secondary,
ES = IS × ZS = 5 V
ES = 4.44 f ∅ N
∴ ∅ = (frac{E_S}{4.44 ,f ,N} = frac{5}{4.44 ,f ,N}) = 45 μWb.

3. A 1000/5 A, 50 Hz correct transformer has a secondary burden comprising a non-inductance of 1.6Ω. The flux in the core at full load is?
a) 160 μWb
b) 180 μWb
c) 200 μWb
d) 150 μWb
Clarification: Turn ratio = 1000/5 = 200
NP = 1
So, NS = 200
Secondary impedance = 1.6Ω
Secondary induced voltage, ES = 5 × 1.6 = 8 V
∴ ES = 4.44 f N ∅
So, ∅ = (frac{E_S}{4.44 ,f ,N} = frac{8}{4.44 ,f ,N}) = 180 μWb.

4. A 1000/5 A, 50 Hz correct transformer has a secondary burden comprising a non-inductance of 2Ω. What is the ratio error, if the iron loss is 3 W and magnetizing mmf is 250 A?
a) 4%
b) 5.7%
c) 3.6%
d) 4.8%
Clarification: E = (frac{10}{200}) V
I = 60 A
I = (frac{mmf}{N_P}) = 250 A
Actual rating R = 200 + (frac{60}{5}) = 212
So, percentage ratio error = (frac{K_n-R}{R}) × 100 = (frac{200-212}{212}) × 100 = 5.7%.

5. A 200/1 Current Transformer (CT) is wound with 400 turns on the secondary on a toroidal core. When it carries a current of 180 A on the primary, the ratio is found to be -0.5%. If the number of secondary turns is reduced by 1, the new ratio error (in %) will be?
a) 0.0
b) -0.5
c) -1.0
d) -2.0
Clarification: Turn compensation only alters ratio error n=400
Ratio error = -0.5% = – (frac{0.5}{100}) × 400 = -2
So, Actual ratio = R = n+1 = 401
Nominal Ratio KN = 400/1 = 400
Now, if the number of turns are reduced by one, n = 399, R = 400
Ratio error = (frac{K_N-R}{R} = frac{200-200}{200}) = 0.

6. A (350 A/ 7A), 50 Hz current transformer has a primary bar. The secondary burden is a pure resistance of 1 Ω and it draws a current of 5 A. The magnetic core requires 350 AT for magnetization. Find the percentage ratio error.
a) 10.56
b) -28.57
c) 11.80
d) -11.80
Clarification: Im = 350/1 = 350 A
Ip = (((nI_s^2)^2 + (I_m^2)^2)^{0.5}) = 490.05
n = 350/7 = 50
∴ R = (frac{I_P}{I_S} = frac{490.05}{7}) = 70
∴ Percentage ratio error = (frac{50-70}{70}) × 100 = -28.57%.

7. The observation when the secondary winding of a current transformer is open-circuited is?
a) The whole of the primary current produces a large value of flux in the core thereby producing a large voltage in the secondary winding
b) The large voltage may act as a safety hazard for the operators and many even raptures the insulation
c) When the large magnetizing force is taken off, it leaves a large value of residual magnetism
d) When the large magnetizing force is taken off, it leaves a small value of residual magnetism
Clarification: Never open the circuit of the secondary winding of a current transformer while to the primary winding is energized. Failure to observe this precaution may lead to serious consequences both to the operating personnel and to the transformer.

8. Usually a CT has ____________
a) Power overload capacity than PT
b) The same overload capacity as a PT
c) A higher overload capacity than a PT
d) No overload capacity
Clarification: A CT cannot have greater or same power overload than PT. Also, it will be having some load on it. This load is a higher overload capacity than a PT.

9. A PT is a device which is ___________
a) Electrostatically coupled
b) Electrically coupled
c) Electromagnetically coupled
d) Conductively coupled