Power systems miscellaneous Difficult Questions and Answers | Page - 39

Power systems miscellaneous

Shunt reactors are sometimes used in high voltage transmission systems to

1. limit the short circuit current through the line
1. compensate for the series reactance of the line under heavily loaded condition
1. limit over-voltages at the load side under lightly loaded condition
1. compensate for the voltage drop in the line under heavily loaded condition

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NA

Correct Option: C

NA

The fuel cost functions of two powr plants are
Plant P₁ : C₁ = 0.05 Pg²₁ + APg₁ + B
Plant P₂ : C₂ = 0.10 Pg²₂ + APg₂ + 2B
Where, P_g1 and P_g2 are the generated powers of two plants, and A and B are the constants. If the two plants optimally share 1000 MW load at increamental fuel cost of 100 Rs/MWh, the ratio of load shared by plants P₁ and P₂ is

1. 1: 4
1. 2: 3
1. 3: 2
1. 4: 1

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C₁ = Pg²₁(0.05) APg₁ + B
Pg₁ + Pg₂ = 1000
C₂ = 0.1Pg²₂ + 3APg₂ + B (dc₁/dPg₂) = 100
Now dc₁/dPg₂ = 2Pg₁ × (0.05) + A
= 2 × 0.05Pg1 + A = 100 ...(i)
and dc₁ × 0.1 Pg₂ + 3A
dPg₂

= 2 × 0.1Pg₂ + 3A = 100 ...(ii)
Solving equation (i) and (ii) we get
Pg₁ = 800 MW
Pg₂ = 200 MW
∴ Pg₁ = 4
Pg₂ 1

Correct Option: D

C₁ = Pg²₁(0.05) APg₁ + B
Pg₁ + Pg₂ = 1000
C₂ = 0.1Pg²₂ + 3APg₂ + B (dc₁/dPg₂) = 100
Now dc₁/dPg₂ = 2Pg₁ × (0.05) + A
= 2 × 0.05Pg1 + A = 100 ...(i)
and dc₁ × 0.1 Pg₂ + 3A
dPg₂

= 2 × 0.1Pg₂ + 3A = 100 ...(ii)
Solving equation (i) and (ii) we get
Pg₁ = 800 MW
Pg₂ = 200 MW
∴ Pg₁ = 4
Pg₂ 1

A two bus power system shown in the figure supplies load of 1.0 + j0.5 p.u.

The values of V₁ in p.u. and δ₂ respectively are

1. 0.95 and 6.00°
1. 1.05 and – 5.44°
1. 1.1 and – 6.00°
1. 1.1 and – 27.12°

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Correct Option: B

A distribution feeder of 1km length having resistance, but negligible reactance, is fed from both the ends by 400V, 50Hz balanced sources. Both voltage sources S₁ and S₁ are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure.

The contributions of S₁ and S₂ in 100A current supplied at location P respectively, are

1. 75A and 25A
1. 50A and 50A
1. 25A and 75A
1. 0A and 100A

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Let I be current supplied by source – 1
‘r’ be resistance/length
400 = (400r) I + (200r) (I – 200) + (200r) (I – 300) + (200r) (I – 500) + 400
∴ 0 = 400rI + 200rI – 40000r + 200rI – 60000r + 200rI – 100000r
∴ 100Ir = 20000r
∴ I = 200A
Current in branch – B is I – 200 = 200 – 200 = 0
4 point p
source – 1 supplies 0A current
source– 2 supplies 100 A current

Correct Option: D

Let I be current supplied by source – 1
‘r’ be resistance/length
400 = (400r) I + (200r) (I – 200) + (200r) (I – 300) + (200r) (I – 500) + 400
∴ 0 = 400rI + 200rI – 40000r + 200rI – 60000r + 200rI – 100000r
∴ 100Ir = 20000r
∴ I = 200A
Current in branch – B is I – 200 = 200 – 200 = 0
4 point p
source – 1 supplies 0A current
source– 2 supplies 100 A current

Power consumed by a balanced 3-phase, 3-wire load is measured by the two wattmeter method. The first wattmeter reads twice that of the second. Then the load impedance angle in radians is

1. π/12
1. π/8
1. π/6
1. π/3

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First wattmeter reads twice that of second when load impedance angle is π/6 radians.

Correct Option: C

First wattmeter reads twice that of second when load impedance angle is π/6 radians.