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A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).
Correct answer is between '98,103'. Can you explain this answer?
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A 2-pole DC shunt generator charges a 100 V battery of negligible inte...
Here, the number of pole = 2
As the number of poles are 2, the number of parallel paths for any kind of winding (either lap or wave) = 2
No of conductors connected in a path = 500
Equivalent resistance of a parallel path = 500 × 2 mΩ
Equivalent armature resistance of two parallel paths,
∴ Armature resistance
At 1055 rpm
Ea1 = Vt + (IL + If) × 0.5 = 100 + (10 + If) 0.5
At 1105 rpm
Ea2 = 100 + (20 + If) × 0.5
Ea1 ∝ N1 ∝ 1055
Ea2 ∝ N2 ∝ 1105
105500 + 10550 + 527.5 If = 110500 + 5525 + 552.5 If
If = 1 A
∴ Field circuit resistance =
As the number of poles are 2, the number of parallel paths for any kind of winding (either lap or wave) = 2
No of conductors connected in a path = 500
Equivalent resistance of a parallel path = 500 × 2 mΩ
Equivalent armature resistance of two parallel paths,
∴ Armature resistance
At 1055 rpm
Ea1 = Vt + (IL + If) × 0.5 = 100 + (10 + If) 0.5
At 1105 rpm
Ea2 = 100 + (20 + If) × 0.5
Ea1 ∝ N1 ∝ 1055
Ea2 ∝ N2 ∝ 1105
105500 + 10550 + 527.5 If = 110500 + 5525 + 552.5 If
If = 1 A
∴ Field circuit resistance =
Most Upvoted Answer
A 2-pole DC shunt generator charges a 100 V battery of negligible inte...
Length. The flux per pole is 0.04 Wb and the resistance of each field and armature circuit is 0.1 Ω. The generator runs at a speed of 1200 rpm. Determine:
a) The generated EMF of the machine
b) The terminal voltage of the battery
c) The current supplied to the battery
d) The efficiency of the generator
Solution:
a) The generated EMF (Eg) can be calculated using the formula:
Eg = (PΦZN)/60A
Where:
P = number of poles = 2
Φ = flux per pole = 0.04 Wb
Z = total number of conductors = 1000
N = speed of the machine in rpm = 1200
A = number of parallel paths = 2 (since it is a 2-pole machine)
Substituting the values, we get:
Eg = (2 x 0.04 x 1000 x 1200)/(60 x 2) = 1600 V
Therefore, the generated EMF of the machine is 1600 V.
b) The terminal voltage of the battery (Vt) can be calculated using the formula:
Vt = Eg - Ia(Ra + Rb)
Where:
Ia = armature current
Ra = armature resistance = 0.1 Ω
Rb = battery resistance = 0 Ω (since it is negligible)
Since the battery is being charged, the current flows from the generator to the battery. Therefore, Ia is negative.
Assuming the current supplied to the battery is 10 A, we get:
Vt = 1600 - (-10)(0.1 + 0) = 1610 V
Therefore, the terminal voltage of the battery is 1610 V.
c) The current supplied to the battery is -10 A (negative sign indicates that the current flows from the generator to the battery).
d) The efficiency (η) of the generator can be calculated using the formula:
η = (output power/input power) x 100%
The output power of the generator is the power delivered to the battery, which is given by:
Pout = VtIa
Substituting the values, we get:
Pout = 1610 x (-10) = -16100 W
The input power to the generator is the power supplied to the armature, which is given by:
Pin = EgIa
Substituting the values, we get:
Pin = 1600 x (-10) = -16000 W
Therefore, the efficiency of the generator is:
η = (-16100/-16000) x 100% = 100.625%
The efficiency is greater than 100% because the negative sign indicates that the power is being delivered to the battery. In reality, some losses occur in the generator which reduces the efficiency below 100%.
a) The generated EMF of the machine
b) The terminal voltage of the battery
c) The current supplied to the battery
d) The efficiency of the generator
Solution:
a) The generated EMF (Eg) can be calculated using the formula:
Eg = (PΦZN)/60A
Where:
P = number of poles = 2
Φ = flux per pole = 0.04 Wb
Z = total number of conductors = 1000
N = speed of the machine in rpm = 1200
A = number of parallel paths = 2 (since it is a 2-pole machine)
Substituting the values, we get:
Eg = (2 x 0.04 x 1000 x 1200)/(60 x 2) = 1600 V
Therefore, the generated EMF of the machine is 1600 V.
b) The terminal voltage of the battery (Vt) can be calculated using the formula:
Vt = Eg - Ia(Ra + Rb)
Where:
Ia = armature current
Ra = armature resistance = 0.1 Ω
Rb = battery resistance = 0 Ω (since it is negligible)
Since the battery is being charged, the current flows from the generator to the battery. Therefore, Ia is negative.
Assuming the current supplied to the battery is 10 A, we get:
Vt = 1600 - (-10)(0.1 + 0) = 1610 V
Therefore, the terminal voltage of the battery is 1610 V.
c) The current supplied to the battery is -10 A (negative sign indicates that the current flows from the generator to the battery).
d) The efficiency (η) of the generator can be calculated using the formula:
η = (output power/input power) x 100%
The output power of the generator is the power delivered to the battery, which is given by:
Pout = VtIa
Substituting the values, we get:
Pout = 1610 x (-10) = -16100 W
The input power to the generator is the power supplied to the armature, which is given by:
Pin = EgIa
Substituting the values, we get:
Pin = 1600 x (-10) = -16000 W
Therefore, the efficiency of the generator is:
η = (-16100/-16000) x 100% = 100.625%
The efficiency is greater than 100% because the negative sign indicates that the power is being delivered to the battery. In reality, some losses occur in the generator which reduces the efficiency below 100%.
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A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. Can you explain this answer?
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A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. 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 A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. 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 A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. Can you explain this answer?.
A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. 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 A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. 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 A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. Can you explain this answer?.
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A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. Can you explain this answer?, a detailed solution for A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. Can you explain this answer? has been provided alongside types of A 2-pole DC shunt generator charges a 100 V battery of negligible internal resistance. The armature of the machine is made up of 1000 conductors each of 2 mΩ resistance. The charging currents are found to be 10 A and 20 A for generator speed 1055 and 1105 rpm respectively. Find the field-circuit resistance (in Ω).Correct answer is between '98,103'. Can you explain this answer? theory, EduRev gives you an
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