Power systems miscellaneous Difficult Questions and Answers | Page - 1

Power systems miscellaneous

If r is radius of the conductor and R is radius of the sheath of strength, then

1. r /R > 1·0
1. r /R > 1·0
1. r/R < 0·632
1. r/R < 0·368

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Gradient at a distance x,
g = V
xIn(R/r)

∴ g_min = V , and
RIn R
r

g_max = V
rIn R
r

and plot of the gradient at the surface of the conductor and the ratio (r/R) is shown in the figure and it can be seen that
r < 1 < 0.368
R e

Correct Option: D

Gradient at a distance x,
g = V
xIn(R/r)

∴ g_min = V , and
RIn R
r

g_max = V
rIn R
r

and plot of the gradient at the surface of the conductor and the ratio (r/R) is shown in the figure and it can be seen that
r < 1 < 0.368
R e

If δ is the loss angle of the cable, then its power factor is

1. sin δ
1. cos δ
1. independent of δ
1. depends upon δ but it is not as per (a) and (b)

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Figure depicts the equivalent circuit for a cable represented by leak age resistance R and capacitance C.
From phasor diagram,
power factor angle, φ = 90° – δ
or cos φ = cos (90° – δ) = sin δ

Correct Option: A

Figure depicts the equivalent circuit for a cable represented by leak age resistance R and capacitance C.
From phasor diagram,
power factor angle, φ = 90° – δ
or cos φ = cos (90° – δ) = sin δ

Three insulating materials with same maximum working stress and permittivities 2·5, 3·0, 4·0, are used in a single core cable, then location of the materials with respect to the core of the cable will be

1. 2·5, 3·0, 4·0
1. 3·0, 2·5, 4·0
1. 4·0, 3·0, 2·5
1. 4·0, 2·5, 3·0

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When all the three materials are subjected to the same maximum stress,

g_max = λ = λ = λ
2πε₁r 2πε₂r₁ 2πε₃r₂

or ε₁r = ε₂r₁ = ε₃r₂
Since r < r₁ < r₂,
Therefore ε₁ > ε₂ > ε₃
Thus, the dielectric material with highest permittivity should be placed near the conductor and other layer in the descending order.

Correct Option: C

When all the three materials are subjected to the same maximum stress,

g_max = λ = λ = λ
2πε₁r 2πε₂r₁ 2πε₃r₂

or ε₁r = ε₂r₁ = ε₃r₂
Since r < r₁ < r₂,
Therefore ε₁ > ε₂ > ε₃
Thus, the dielectric material with highest permittivity should be placed near the conductor and other layer in the descending order.

For the same voltage boost, the reactive power capacity is

1. more for shunt capacitor
1. more for series capacitor
1. it is same for both series and shunt
1. uncertain

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Let Q'_c be t he reactive power of the shunt capacitor, E_r receiving end voltage, X is the line reactance.
Then voltage drop across it
= Q_c . X
E_r

where, Q_c be the reactive power of the series capacitor, then voltage drop across it
sin φ_v = Q_c ,
I

where φ_v is the power factor angle.
For equal voltage boost,
Q'_c X = Q_csin φ_v
E_r I

or Q'_c = sin φ_v = 0.6 = 6
Q_c IX/E_r 0.1

Q'_c > Q_c

Correct Option: A

Let Q'_c be t he reactive power of the shunt capacitor, E_r receiving end voltage, X is the line reactance.
Then voltage drop across it
= Q_c . X
E_r

where, Q_c be the reactive power of the series capacitor, then voltage drop across it
sin φ_v = Q_c ,
I

where φ_v is the power factor angle.
For equal voltage boost,
Q'_c X = Q_csin φ_v
E_r I

or Q'_c = sin φ_v = 0.6 = 6
Q_c IX/E_r 0.1

Q'_c > Q_c

If corona loss on a particular system at 50 Hz is 1 kW/phase per km, then corona loss on the same system with supply frequency 25 Hz will be

1. 1 kW/phase/km
1. 0·5 kW/phase/km
1. 0·667 kW/phase/km
1. none of these

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Corona loss
(P) = 241 × 10^–5 (ƒ + 25) √ r (V_p - V₀)²,
δ d

where, ƒ = supply frequency
nOW P ∝ ƒ + 25
∴ P₁ = ƒ₁ + 25
P₂ ƒ₂ + 25

or P₂ = ƒ₂ + 25
ƒ₁ + 25

or P₁ = 50 × 1kW = /b>0.667kW/phase/km
75

Correct Option: C

Corona loss
(P) = 241 × 10^–5 (ƒ + 25) √ r (V_p - V₀)²,
δ d

where, ƒ = supply frequency
nOW P ∝ ƒ + 25
∴ P₁ = ƒ₁ + 25
P₂ ƒ₂ + 25

or P₂ = ƒ₂ + 25
ƒ₁ + 25

or P₁ = 50 × 1kW = /b>0.667kW/phase/km
75