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A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added is
- a)4.5 Ω
- b)3 Ω
- c)2.5 Ω
- d)2 Ω
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A 200V dc series motor develops its rated output at 1500 rpm while ta...
⊤ ∝ la2 ⇒ T = kla2
For const torque armature I must be const.
Eb = 200 − 20 × (0.6 + 0.4) = 180V
la = 20A for 1000rpm. Ea = kϕw
180Ea = 1500 / 1000
Ea = 120V = 200 − 20(0.4 + 0.6 + Rext)
⇒ Rext = 200 − 120 / 20 −1 = 3Ω
Most Upvoted Answer
A 200V dc series motor develops its rated output at 1500 rpm while ta...
Solution:
Given data:
Voltage (V) = 200 V
Speed (N1) = 1500 rpm
Armature current (Ia) = 20 A
Armature resistance (Ra) = 0.6 Ω
Series field resistance (Rs) = 0.4 Ω
To find:
External resistance required to obtain rated torque at 1000 rpm.
Formula used:
T ∝ Φ Ia
Φ = (V - Ia Ra) / N
T = K Φ Ia
Where,
T = torque
Φ = flux
Ia = armature current
V = applied voltage
Ra = armature resistance
N = speed
K = constant
Calculation:
Given, K is constant for the motor.
Therefore, T1 = K Φ Ia1 and T2 = K Φ Ia2,
where T1 and Ia1 correspond to the rated output at 1500 rpm,
and T2 and Ia2 correspond to the rated torque at 1000 rpm.
At rated output:
T1 = 1.2 × V × Ia1 / N1 [From the torque equation]
T1 = K Φ Ia1 [From the flux equation]
∴ Ia1 = T1 / K Φ
Φ = (V - Ia1 Ra) / N1
∴ Φ = (200 - 20 × 0.6) / 1500
Φ = 0.11 Wb
At rated torque:
T2 = 1.2 × V × Ia2 / N2 [From the torque equation]
T2 = K Φ Ia2 [From the flux equation]
∴ Ia2 = T2 / K Φ
Φ = (V - Ia2 Ra) / N2
To obtain T2 at N2 = 1000 rpm,
we need to add external resistance (Re) in series with the armature circuit.
Let the value of Re be R.
∴ Ia2 = (V - Ia2 (Ra + R)) / N2
T2 = K Φ Ia2
∴ T2 = K Φ (V - Ia2 (Ra + R)) / N2
∴ T2 = K Φ (V - T2 / K Φ (Ra + R)) / N2
∴ T2 = (K Φ V N2) / (K Φ N2 (Ra + R) + V)
∴ T2 = 1.2 × V × Ia2 / N2
∴ Ia2 = T2 / (1.2 × Φ × N2)
∴ T2 / (1.2 × Φ × N2) = (V - T2 / (1.2 × Φ × N2) (Ra + R)) / N2
∴ T2 = K Φ (V - Ia2 (Ra + R)) / N2
∴ T2 = K Φ (V - T2 / K Φ (Ra + R)) / N2
∴ T2 = (K Φ V N2) / (K Φ N2 (Ra + R) + V)
Substit
Given data:
Voltage (V) = 200 V
Speed (N1) = 1500 rpm
Armature current (Ia) = 20 A
Armature resistance (Ra) = 0.6 Ω
Series field resistance (Rs) = 0.4 Ω
To find:
External resistance required to obtain rated torque at 1000 rpm.
Formula used:
T ∝ Φ Ia
Φ = (V - Ia Ra) / N
T = K Φ Ia
Where,
T = torque
Φ = flux
Ia = armature current
V = applied voltage
Ra = armature resistance
N = speed
K = constant
Calculation:
Given, K is constant for the motor.
Therefore, T1 = K Φ Ia1 and T2 = K Φ Ia2,
where T1 and Ia1 correspond to the rated output at 1500 rpm,
and T2 and Ia2 correspond to the rated torque at 1000 rpm.
At rated output:
T1 = 1.2 × V × Ia1 / N1 [From the torque equation]
T1 = K Φ Ia1 [From the flux equation]
∴ Ia1 = T1 / K Φ
Φ = (V - Ia1 Ra) / N1
∴ Φ = (200 - 20 × 0.6) / 1500
Φ = 0.11 Wb
At rated torque:
T2 = 1.2 × V × Ia2 / N2 [From the torque equation]
T2 = K Φ Ia2 [From the flux equation]
∴ Ia2 = T2 / K Φ
Φ = (V - Ia2 Ra) / N2
To obtain T2 at N2 = 1000 rpm,
we need to add external resistance (Re) in series with the armature circuit.
Let the value of Re be R.
∴ Ia2 = (V - Ia2 (Ra + R)) / N2
T2 = K Φ Ia2
∴ T2 = K Φ (V - Ia2 (Ra + R)) / N2
∴ T2 = K Φ (V - T2 / K Φ (Ra + R)) / N2
∴ T2 = (K Φ V N2) / (K Φ N2 (Ra + R) + V)
∴ T2 = 1.2 × V × Ia2 / N2
∴ Ia2 = T2 / (1.2 × Φ × N2)
∴ T2 / (1.2 × Φ × N2) = (V - T2 / (1.2 × Φ × N2) (Ra + R)) / N2
∴ T2 = K Φ (V - Ia2 (Ra + R)) / N2
∴ T2 = K Φ (V - T2 / K Φ (Ra + R)) / N2
∴ T2 = (K Φ V N2) / (K Φ N2 (Ra + R) + V)
Substit
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A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer?
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A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer?.
A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer?.
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A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added isa)4.5 Ωb)3 Ωc)2.5 Ωd)2 ΩCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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