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In a two-wire system, the voltage across the supply end is maintained at 500 V. The line is 4 km long. If the full-load current is 15 A, what should be the booster voltage and output so that the distant voltage can also be 500 V?
Take the resistance of the cable to be 0.5 ohm/km.
  • a)
    224 W
  • b)
    260 W
  • c)
    450 W
  • d)
    120 W
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
In a two-wire system, the voltage across the supply end is maintained ...
Given data:
- Voltage across the supply end (V1): 500 V
- Length of the line (L): 4 km
- Full-load current (I): 15 A
- Resistance of the cable (R): 0.5 ohm/km

To maintain a distant voltage of 500 V, we need to compensate for the voltage drop caused by the resistance of the cable. This can be achieved by using a booster at the supply end.

Let's calculate the voltage drop first using Ohm's law:
Voltage drop (Vd) = I * R * L

Substituting the given values:
Vd = 15 A * 0.5 ohm/km * 4 km
Vd = 30 V

To maintain the distant voltage at 500 V, we need to add this voltage drop to the supply voltage:
V2 = V1 + Vd
V2 = 500 V + 30 V
V2 = 530 V

Now, let's calculate the booster voltage required:
Booster voltage (Vb) = V2 - V1
Vb = 530 V - 500 V
Vb = 30 V

To calculate the output power, we need to consider the current flowing through the booster. Since the booster is placed at the supply end, the current flowing through it will be the same as the full-load current.

Output power (P) = Vb * I
P = 30 V * 15 A
P = 450 W

Therefore, the booster voltage required is 30 V and the output power is 450 W. Hence, the correct answer is option C.
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Community Answer
In a two-wire system, the voltage across the supply end is maintained ...
We Have,
Resistance of cable is 0.5 ohm/km
Hence, total Resistance of 4 km long cable (R) = 0.5 × 4 = 2 Ω
Load Current (I) = 15 A
So total Voltage drop in the cable = IR = 15 × 2 = 30 volts
Total Power Loss (P) = I2R
⇒ P = I2R = 152 × 2 = 225 × 2 = 450 W
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