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A 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)
Correct answer is '260'. Can you explain this answer?
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A 250 V DC shunt generator delivers a load current of 195 A. The arma...
Solution:
Given:
Voltage of the generator (V) = 250 V
Load current (I) = 195 A
Armature resistance (Ra) = 0.05 Ω
Shunt field resistance (Rf) = 50 Ω
To find:
Generated electromotive force (Ea) in the armature
Using Ohm's law, we can find the voltage drop across the armature resistance and the shunt field resistance.
1. Voltage drop across the armature resistance:
Voltage drop (Va) = I * Ra
Substituting the given values:
Va = 195 A * 0.05 Ω
Va = 9.75 V
2. Voltage drop across the shunt field resistance:
Voltage drop (Vf) = I * Rf
Substituting the given values:
Vf = 195 A * 50 Ω
Vf = 9750 V
The total generated electromotive force (Eg) can be calculated by adding the voltage drops across the armature and shunt field resistances.
3. Total generated electromotive force:
Eg = V + Va + Vf
Substituting the given values:
Eg = 250 V + 9.75 V + 9750 V
Eg = 10,009.75 V
However, the question asks for the value of the generated electromotive force in the armature (Ea), so we need to subtract the voltage drop across the shunt field resistance from the total generated electromotive force.
4. Generated electromotive force in the armature:
Ea = Eg - Vf
Substituting the given values:
Ea = 10,009.75 V - 9750 V
Ea = 259.75 V
Rounding off to the nearest integer, the value of the generated electromotive force in the armature is 260 V, which matches the given correct answer.
Given:
Voltage of the generator (V) = 250 V
Load current (I) = 195 A
Armature resistance (Ra) = 0.05 Ω
Shunt field resistance (Rf) = 50 Ω
To find:
Generated electromotive force (Ea) in the armature
Using Ohm's law, we can find the voltage drop across the armature resistance and the shunt field resistance.
1. Voltage drop across the armature resistance:
Voltage drop (Va) = I * Ra
Substituting the given values:
Va = 195 A * 0.05 Ω
Va = 9.75 V
2. Voltage drop across the shunt field resistance:
Voltage drop (Vf) = I * Rf
Substituting the given values:
Vf = 195 A * 50 Ω
Vf = 9750 V
The total generated electromotive force (Eg) can be calculated by adding the voltage drops across the armature and shunt field resistances.
3. Total generated electromotive force:
Eg = V + Va + Vf
Substituting the given values:
Eg = 250 V + 9.75 V + 9750 V
Eg = 10,009.75 V
However, the question asks for the value of the generated electromotive force in the armature (Ea), so we need to subtract the voltage drop across the shunt field resistance from the total generated electromotive force.
4. Generated electromotive force in the armature:
Ea = Eg - Vf
Substituting the given values:
Ea = 10,009.75 V - 9750 V
Ea = 259.75 V
Rounding off to the nearest integer, the value of the generated electromotive force in the armature is 260 V, which matches the given correct answer.
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A 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. Can you explain this answer?
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A 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. 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 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. 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 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. Can you explain this answer?.
A 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. 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 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. 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 250 V DC shunt generator delivers a load current of 195 A. The armature and shunt field resistances are 0.05 Ω and 50 Ω, respectively. What is the value of generated e.m.f. (in volts) in the armature? (Answer up to the nearest integer)Correct answer is '260'. Can you explain this answer?.
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