1. A transmission line has a reactance of 4 Pu is operating at Vs = Vr = 2 Pu. The generator is connected at the source end which is delivering 0.5 Pu of active power. Find the load angle?
A. 60°
B. 45°
C. 30°
D. 35°
Answer With Explanation:
C. 30°
Solution:
Power delivered P = (Vs × Vr × sinδ)/X
Vs = sending end voltage = 2
Vr = receiving end voltage = 2
X = reactance = 4
P = 0.5 pu
δ = load angle
0.5 = (2 × 2× sinδ)/4
sin-1(0.5) = δ
δ = 30°
2. Find the number of strands of ACSR conductor for 3 layer transmission line?
A. 10
B. 19
C. 29
D. 27
Answer with Explanation:
B. 19
Solution:
The total number of strands in ACSR conductor is given as
N = 3x² – 3x + 1 where
x = number of layer
So the total number of strands
N = 3×3² – 3×3 + 1 = 19
3. Find the total diameter of the ACSR conductor with 2 layers and the diameter of each strand is 3?
A. 9B. 10
C. 5
D. 15
Answer with Explanation:
A. 9
Solution:
The diameter of the ACSR conductor can be calculated by
D = (2x – 1)xd where
x = number of layer
d = diameter of each strand
Therefore, Total diameter of ACSR conductor
D = (2×2 -1)3
D = 9
4. Four identical alternators each are rated for 20 MVA, 11 KV having a subtransient reactance of 16% are working in parallel. The short circuit level at the busbar is
A. 500 MVAB. 400 MVA
C. 125 MVA
D. 100 MVA
Answer With Explanation:
A. 500 MVA
Solution:
5.If the fault current is 2000 A, the relay setting is 50% and CT ratio is 400 : 5, then plug setting multiplier will be
A. 10.
B. 15.
C. 25.
D. 50.
Answer with Explanation:
A. 10.
Solution:
The current transformer ratio is 400 / 5.
So, the CT can be provided a rated current of 5 Amps to the relay coil. Thereby the rated relay current is 5 Amps.
If the relay setting is 50%, then the relay can be operated for 5 × ( 50 / 100 ) = 2.5 Amps.
This is the pick-up current of that relay. We know that the fault current is 2000 Amps.
Hence, the fault current in the secondary of the CT is 5 × ( 2000 / 400) = 25 Amps. Hence, the plug setting multiplier = 25 / 2.5 = 10.
That means the fault current in the secondary of the CT is ten times greater than the operating current of the relay coil.
6. A transmission line has a reactance of 1 Pu is operating at Vs = Vr = 1 Pu. The generator is connected at source end which is delivering 0.5 Pu of active power and the transmission line is compensated with a series capacitance of 0.5 Pu. Find the load angle with series capacitance compensation?
A. 14.5°
B. 29°
C. 35.5°
D. 10.5°
Answer With Explanation:
A. 14.5°
Solution:
With series capacitance compensation power received
Vs = sending end voltage
Vr = receiving end voltage
δ = load angle
7. In a 4*4 Y bus matrix the number of non zero elements are 12. Find the sparsity of the system?
A. 0.25
B. 0.5
C. 0.75
D. 1
A. 0.25
Solution:
Total number of elements in 4×4 Y bus matrix = 16
Number of zero elements = 4
8. In a 4 bus,4×4 Y bus matrix the number of non zero elements are 12. Find the number of transmission lines?
A. 8
B. 4
C. 2
D. 5
Answer with Explanation:
B. 4
Solution:
N = Number of buses
X = Sparsity
Sparsity X = 4/16 = 0.25
9. A network containing 100 buses in which 10 are the voltage control buses, 5 are fixed shunt capacitor buses, 20 are the reactive power support buses, 6 are the generator buses. Find the size of the Jacobian matrix?
A. 163 × 163
B. 164 × 164
C. 165 × 165
D. 162 × 162
Answer with Explanation:
A. 163 × 163
Solution:
Size of the Jacobian matrix = (2n-m-2)×(2n-m-2)
Where,
n = Total number of buses
m = Total number of PV buses
m = Voltage control buses + Reactive power support buses + generator buses except slack bus.
Fixed shunt capacitors are supplying a constant amount of reactive power, so that fixed shunt capacitors are considered as load buses or PQ buses.
Therefore, n = 100 and m =35
Size of the Jacobian matrix = (2×100 - 35 -2)×(2×100 - 35 - 2) = 163×163
10. A network containing 50 buses in which 10 are the voltage control buses, 6 are the generator buses. Find the size of the Jacobian matrix?
A. 84*84
B. 83*83
C. 34*34
D. 33*33
Answer With Explanation:
B. 83*83
Solution:
Size of the Jacobian matrix = (2n-m-2)*(2n-m-2)
Where,
n = Total number of buses
m = Total number of PV buses
m = Voltage control buses + Reactive power support buses + generator buses except slack bus
Size of the Jacobian matrix = (2*50 -15 - 2)*(2*50 - 15 -) =83*83
11. If the sparsity of a 5 bus transmission line is 0.4. Find the number of transmission lines?
A. 6
B. 5
C. 4
D. 3
Answer with Explanation:
B. 5
Solution:
n = number of buses
x = sparsity
12. A 4 bus power system consists of 4 transmission lines, then find the sparsity of Y bus matrix?
A. 0.25
B. 0.5
C. 0.75
D. 0.4
Answer With Explanation:
A. 0.25
Solution:
Where,
n = number of buses
x = sparsity
13. Find the number of strands of ACSR conductor for 4 layer transmission line?
A. 1
B. 7
C. 37
D. 29
Answer with Explanation:
C. 37
Solution:
Total number of strands N = 3x² - 3x + 1
Where, x = Number of layers = 4
Total number of strands N = 3×4² - 3×4 + 1 = 37
Electrical Facts:
