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A 400 V, three-phase, four-wire is feeding a balanced load of 100 W lamp in each phase. If the lamp in one of the phases goes off. What will be the reading of the ammeter connecting neutrals of the supply and load?
- a)0.43 A
- b)0.86 A
- c)4.61 A
- d)2.31 A
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A 400 V, three-phase, four-wire is feeding a balanced load of 100 W l...
During balance load total power P = 3 × 100 = 300 W
View all questions of this test
If the lamp in one of the phases goes off, then total power = 300 × 0.57 = 171 W
∴ the reading of ammeter connecting neutrals = 171/400 = 0.43 A
Most Upvoted Answer
A 400 V, three-phase, four-wire is feeding a balanced load of 100 W l...
Given information:
- Voltage: 400 V
- Three-phase, four-wire system
- Balanced load: 100 W lamp in each phase
To find: Reading of the ammeter connecting neutrals of the supply and load when one of the lamps goes off.
Explanation:
1. Balanced Load:
In a balanced load, the three phases are identical and draw equal power. Since each phase has a 100 W lamp, the total power consumed by the load is:
Total Power = Power per Phase x Number of Phases
Total Power = 100 W x 3 = 300 W
2. Neutral Current in Balanced Load:
In a balanced load, the neutral current is ideally zero. This is because the currents in the three phases cancel each other out in the neutral conductor.
3. Lamp Goes Off:
When one of the lamps goes off, the load becomes unbalanced. The two remaining lamps continue to draw power, but the faulty lamp consumes no power.
4. Unbalanced Load:
In an unbalanced load, the currents in the three phases are not equal, resulting in a non-zero neutral current.
5. Calculation:
Since the faulty lamp is not consuming any power, the total power consumed by the load reduces to:
Total Power = Power per Phase x Number of Remaining Phases
Total Power = 100 W x 2 = 200 W
To calculate the neutral current, we can use the formula:
Neutral Current = Total Power / Line-to-Neutral Voltage
Neutral Current = 200 W / 400 V
Neutral Current = 0.5 A
However, the question asks for the reading of the ammeter connecting neutrals of the supply and load. In a balanced load, the neutral current is ideally zero. When one of the lamps goes off, the load becomes unbalanced and a non-zero neutral current flows. Therefore, the reading of the ammeter will be 0.5 A, which is option A.
- Voltage: 400 V
- Three-phase, four-wire system
- Balanced load: 100 W lamp in each phase
To find: Reading of the ammeter connecting neutrals of the supply and load when one of the lamps goes off.
Explanation:
1. Balanced Load:
In a balanced load, the three phases are identical and draw equal power. Since each phase has a 100 W lamp, the total power consumed by the load is:
Total Power = Power per Phase x Number of Phases
Total Power = 100 W x 3 = 300 W
2. Neutral Current in Balanced Load:
In a balanced load, the neutral current is ideally zero. This is because the currents in the three phases cancel each other out in the neutral conductor.
3. Lamp Goes Off:
When one of the lamps goes off, the load becomes unbalanced. The two remaining lamps continue to draw power, but the faulty lamp consumes no power.
4. Unbalanced Load:
In an unbalanced load, the currents in the three phases are not equal, resulting in a non-zero neutral current.
5. Calculation:
Since the faulty lamp is not consuming any power, the total power consumed by the load reduces to:
Total Power = Power per Phase x Number of Remaining Phases
Total Power = 100 W x 2 = 200 W
To calculate the neutral current, we can use the formula:
Neutral Current = Total Power / Line-to-Neutral Voltage
Neutral Current = 200 W / 400 V
Neutral Current = 0.5 A
However, the question asks for the reading of the ammeter connecting neutrals of the supply and load. In a balanced load, the neutral current is ideally zero. When one of the lamps goes off, the load becomes unbalanced and a non-zero neutral current flows. Therefore, the reading of the ammeter will be 0.5 A, which is option A.
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A 400 V, three-phase, four-wire is feeding a balanced load of 100 W lamp in each phase. If the lamp in one of the phases goes off. What will be the reading of the ammeter connecting neutrals of the supply and load?a)0.43 Ab)0.86 Ac)4.61 Ad)2.31 ACorrect answer is option 'A'. Can you explain this answer?
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