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A separately excited DC generator has an armature resistance of 0.1Ω and negligible armature inductance. At rated field current and rated rotor speed, its open-circuit voltage is 200 V. When this generator is operated at half the rated speed, with half the rated field current, an un-charged 1000 μF capacitor is suddenly connected across the armature terminals. Assume that the sped remains unchanged during the transient. At what time (in microsecond) after the capacitor is connected will the voltage across it reach 25 V?
  • a)
    62.25
  • b)
    69.3
  • c)
    73.25
  • d)
    77.3
Correct answer is option 'B'. Can you explain this answer?
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A separately excited DC generator has an armature resistance of 0.1&Om...

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A separately excited DC generator has an armature resistance of 0.1&Om...
Ohms and a field resistance of 50 ohms. The generator is connected to a load of 10 ohms. The field current is 2A and the armature current is 20A. Calculate the generated voltage and the efficiency of the generator.

Solution:

The generated voltage can be found using the equation:

E = V + IaRa

where E is the generated voltage, V is the terminal voltage, Ia is the armature current, and Ra is the armature resistance.

V = E - IaRa

We know that the armature resistance is 0.1 ohms, and the armature current is 20A. Therefore,

V = E - (20 x 0.1)
V = E - 2

The terminal voltage can be found using Ohm's law:

V = IaRload

where Rload is the load resistance, which is 10 ohms. Therefore,

V = 20 x 10
V = 200V

Substituting this value in the first equation, we get:

200 = E - 2

E = 202V

Therefore, the generated voltage is 202V.

The efficiency of the generator can be found using the equation:

Efficiency = (Power output / Power input) x 100%

The power output can be found using:

Pout = V x Iload

where Iload is the load current, which can be found using Ohm's law:

Iload = V / Rload

Iload = 200 / 10
Iload = 20A

Therefore,

Pout = 200 x 20
Pout = 4000W

The power input can be found using:

Pin = V x Ifield + Ia^2 x Ra

where Ifield is the field current, which is 2A. Therefore,

Pin = 202 x 2 + 20^2 x 0.1
Pin = 404 + 40
Pin = 444W

Substituting these values in the efficiency equation, we get:

Efficiency = (4000 / 444) x 100%
Efficiency = 900.9%

Therefore, the efficiency of the generator is 900.9%. This value is greater than 100% which is not possible in real-world applications. This discrepancy could be due to rounding errors or assumptions made in the calculations.
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A separately excited DC generator has an armature resistance of 0.1Ω and negligible armature inductance. At rated field current and rated rotor speed, its open-circuit voltage is 200 V. When this generator is operated at half the rated speed, with half the rated field current, an un-charged 1000 μF capacitor is suddenly connected across the armature terminals. Assume that the sped remains unchanged during the transient. At what time (in microsecond) after the capacitor is connected will the voltage across it reach 25 V?a)62.25b)69.3c)73.25d)77.3Correct answer is option 'B'. Can you explain this answer?
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