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An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)
Correct answer is between '5.4,5.6'. Can you explain this answer?
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An electric train taking of constant current of 800A moves on a sectio...
Potential of P, Vp = 600 – I1 (0.05)x
= 640 – (800 – I1) (0.05) (10 – x)
⇒ (800 – I1) (0.05) (10 – x) = 40 + I1 (0.05) x
⇒ (800 – I1) (10 – x) = 800 + I1 x
⇒ 800 + I1 x – 10I1 – 800x = 800 + I1 x
⇒ 100I1 + 800x = 7200
⇒ I1 = 720 – 80 x
Now, Vp = 600 – (720 – 80x) 0.05 x
= 600 – 360x + 4x2
For Vp to be minimum
⇒ -36 + 8x = 0
⇒ x = 4.5 km
Distance from 640V substation is = 10 – x = 5.5 km
Most Upvoted Answer
An electric train taking of constant current of 800A moves on a sectio...
To calculate the power consumed by the electric train, we can use the formula:
Power = Current^2 * Resistance
Given:
Current = 800A
Resistance = 0.05 Ω
First, we need to calculate the voltage drop across the track due to its resistance. We can use Ohm's Law:
Voltage drop = Current * Resistance
Voltage drop = 800A * 0.05 Ω
Voltage drop = 40V
Next, we need to calculate the effective voltage across the train. This can be done by subtracting the voltage drop from the higher voltage of the substation:
Effective voltage = Higher voltage - Voltage drop
Effective voltage = 640V - 40V
Effective voltage = 600V
Now, we can calculate the power consumed by the electric train using the effective voltage and the given current:
Power = Current^2 * Resistance
Power = 800A^2 * 0.05 Ω
Power = 640,000W
Therefore, the electric train consumes 640,000 watts of power.
Power = Current^2 * Resistance
Given:
Current = 800A
Resistance = 0.05 Ω
First, we need to calculate the voltage drop across the track due to its resistance. We can use Ohm's Law:
Voltage drop = Current * Resistance
Voltage drop = 800A * 0.05 Ω
Voltage drop = 40V
Next, we need to calculate the effective voltage across the train. This can be done by subtracting the voltage drop from the higher voltage of the substation:
Effective voltage = Higher voltage - Voltage drop
Effective voltage = 640V - 40V
Effective voltage = 600V
Now, we can calculate the power consumed by the electric train using the effective voltage and the given current:
Power = Current^2 * Resistance
Power = 800A^2 * 0.05 Ω
Power = 640,000W
Therefore, the electric train consumes 640,000 watts of power.
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An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer?
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An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer?.
An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electric train taking of constant current of 800A moves on a section of line between two substations 10 km apart maintained at 600 and 640 V respectively the track resistance is 0.05 Ω/km both go and return. The point of minimum potential along the track is at a distance from 640 V substation is – (in km)Correct answer is between '5.4,5.6'. Can you explain this answer?.
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