Power systems miscellaneous
- A single-phase load is supplied by a single-phase volt age source. If the current flowing from the load to the source is 10∠– 150°A and if the voltage at the load terminals is 100 ∠60°V, then the
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V = 100 ∠60°
I = – 10 ∠– 150°C (source to load current)
P complex power
= VI* =100∠60° 10°– 30° = 1000 ∠30°
= 1000 cos 30° + j 1000 sin 30°
Hence load absorbs both active and reactive power.Correct Option: B
V = 100 ∠60°
I = – 10 ∠– 150°C (source to load current)
P complex power
= VI* =100∠60° 10°– 30° = 1000 ∠30°
= 1000 cos 30° + j 1000 sin 30°
Hence load absorbs both active and reactive power.
- A source vs (t) =V cos 100πt has an internal impedance of (4 + j3)Ω If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in Ω should be
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For max power RL = √R²s + X²s = √4² + 3² = 5
Correct Option: C
For max power RL = √R²s + X²s = √4² + 3² = 5
- For a fully transposed transmission line
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Where Zn → neutral impedance
ZS → Self impedance
Zm → mutual impedance
Let Z0 → zero sequence impedance
Z1 → +ve sequence impedance
Z2 → – ve sequence impedanceZ012 = Z000 0Z10 00Z2
∴ Z1 = Z2 = ZS – ZmCorrect Option: B
Where Zn → neutral impedance
ZS → Self impedance
Zm → mutual impedance
Let Z0 → zero sequence impedance
Z1 → +ve sequence impedance
Z2 → – ve sequence impedanceZ012 = Z000 0Z10 00Z2
∴ Z1 = Z2 = ZS – Zm
- In a long transmission line with r, l, g and c are the resistance, inductance, shunt conductance and capacitance per unit length, respectively, the condition for distortionless transmission is
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For distortionless transmission line,
R = G RC = GL L C Correct Option: A
For distortionless transmission line,
R = G RC = GL L C
- The horizontally placed conductors of a single phase line operating at 50 Hz are having outside diameter of 1.6 cm, and the spacing between centers of the conductors is 6 m. The permittivity of free space is 8.854 × 10–12 F/m. The capacitance to ground per kilometer of each line is
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Capacitance to ground/m of each line
C = 2π∈0 F/m In D r
Where D is spacing between conductors and r is the radius of conductor.⇒ C = 2π × 8.854 × 10-12 F/m In 600 0.8
= 8.4 × 10-12 F/m
⇒ C = 8.4 × 10-9 F/kmCorrect Option: B
Capacitance to ground/m of each line
C = 2π∈0 F/m In D r
Where D is spacing between conductors and r is the radius of conductor.⇒ C = 2π × 8.854 × 10-12 F/m In 600 0.8
= 8.4 × 10-12 F/m
⇒ C = 8.4 × 10-9 F/km