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The maximum efficiency occurs in a separately excited d.c. generator when the terminal voltageis 220 V and the induced emf is 240 V, the stray losses, if the armature resistance is 0.2 W , will be
- a)1000 W
- b)2000 W
- c)3000 W
- d)4000 W
Correct answer is option 'B'. Can you explain this answer?
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The maximum efficiency occurs in a separately excited d.c. generator w...
Theory:
The efficiency of a separately excited DC generator can be calculated using the formula:
Efficiency = (Power output / Power input) * 100
The power output is given by the product of terminal voltage and load current, while the power input is given by the product of induced EMF and armature current.
The efficiency is maximum when the power output is equal to the power input. In other words:
Power output = Power input
Therefore, we can equate the expressions for power output and power input and solve for the stray losses.
Solution:
Given:
Terminal voltage (Vt) = 220 V
Induced EMF (E) = 240 V
Armature resistance (Ra) = 0.2 Ω
We know that the power output is given by:
Power output = Vt * Ia
The power input is given by:
Power input = E * Ia
Since the efficiency is maximum, we can equate the power output and power input:
Vt * Ia = E * Ia
Cancelling out Ia from both sides, we get:
Vt = E
Substituting the given values, we have:
220 V = 240 V
This implies that there is a voltage drop of 20 V across the armature resistance due to stray losses.
The stray losses can be calculated using the formula:
Stray losses = (Voltage drop across armature resistance) * (Load current)
Given that the armature resistance is 0.2 Ω, we can calculate the load current as:
Load current = Vt / Ra = 220 V / 0.2 Ω = 1100 A
Substituting the values into the formula, we get:
Stray losses = 20 V * 1100 A = 22000 W = 22000 W
Therefore, the stray losses will be 22000 W, which is equal to 2000 W (option B).
Conclusion:
The stray losses in the separately excited DC generator will be 2000 W.
The efficiency of a separately excited DC generator can be calculated using the formula:
Efficiency = (Power output / Power input) * 100
The power output is given by the product of terminal voltage and load current, while the power input is given by the product of induced EMF and armature current.
The efficiency is maximum when the power output is equal to the power input. In other words:
Power output = Power input
Therefore, we can equate the expressions for power output and power input and solve for the stray losses.
Solution:
Given:
Terminal voltage (Vt) = 220 V
Induced EMF (E) = 240 V
Armature resistance (Ra) = 0.2 Ω
We know that the power output is given by:
Power output = Vt * Ia
The power input is given by:
Power input = E * Ia
Since the efficiency is maximum, we can equate the power output and power input:
Vt * Ia = E * Ia
Cancelling out Ia from both sides, we get:
Vt = E
Substituting the given values, we have:
220 V = 240 V
This implies that there is a voltage drop of 20 V across the armature resistance due to stray losses.
The stray losses can be calculated using the formula:
Stray losses = (Voltage drop across armature resistance) * (Load current)
Given that the armature resistance is 0.2 Ω, we can calculate the load current as:
Load current = Vt / Ra = 220 V / 0.2 Ω = 1100 A
Substituting the values into the formula, we get:
Stray losses = 20 V * 1100 A = 22000 W = 22000 W
Therefore, the stray losses will be 22000 W, which is equal to 2000 W (option B).
Conclusion:
The stray losses in the separately excited DC generator will be 2000 W.
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