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°
Answer: a
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. I_e = 0. I_M = 300 A.
Secondary winding current I_S = 7 A
Reflected secondary winding current = n I_S = 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
Answer: b
Clarification: Voltage induced in the secondary,
E_S = I_S × Z_S = 5 V
E_S = 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
Answer: b
Clarification: Turn ratio = 1000/5 = 200
N_P = 1
So, N_S = 200
Secondary impedance = 1.6Ω
Secondary induced voltage, E_S = 5 × 1.6 = 8 V
∴ E_S = 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%
Answer: b
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
Answer: c
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 K_N = 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
Answer: b
Clarification: I_m = 350/1 = 350 A
I_p = (((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
Answer: b
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
Answer: c
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
Answer: c
Clarification: A Potential Transformer cannot be electrostatically coupled since CRO are electrostatically coupled. Also, they cannot be conductively coupled. But since they are kind of electrically coupled hence electromagnetically coupled is the only correct option.

10. The CT supplies current to the current coil of a wattmeter power factor meter, energy meter and, an ammeter. These are connected as?
a) All coils in parallel
b) All coils in series
c) Series-parallel connection with two in each arm
d) Series-parallel connection with one in each arm
Answer: b
Clarification: Since the CT supplies the current to the current coil of a wattmeter, therefore the coils are connected in series so that the current remains the same. If they were connected in parallel then the voltages would have been the same but the currents would not be the same and thus efficiency would decrease.

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