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A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r.p.m. The resistance of shunt field is 50Ω and that of armature 0.1Ω The iron and friction losses amount to 1600W. What is the shaft torque?
  • a)
    152.6 N-m
  • b)
    158.6 N-m
  • c)
    166.4 N-m
  • d)
    172.6 N-m
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r...
Since Ish = 220/ 50 = 4.4 A
Ia = 80-4.4 = 75.6 Amp
Eb= V - IaRa
= 220- (75.6 x 0.1) = 212.44
Driving Power Eb.Ia = 212.44x75.6
= 16060 N-m
Output Power = Driving Power -  Iron and Friction Losses
= 16060-1600
= 14460 W
Armature Torque
Ta = EbIa/2πN   if N in r.p.s
Ta = (60EbIa)/(2πN)  if N in r.p.m
so; Ta = (9.55EbIa)/N   if N in r.p.s
= (9.55 x 212.44 x 75.6)/800
= 191.7 N-m
Since the whole of armature torque is not available for doing useful work, a certain part is required to overcome iron & friction losses
Hence Lost torque = 9.55x (1600/800)
=  19.1 N-m
Hence Useful Shaft torque = Ta - Tlost
= 191.7 - 19.1
= 172.6 N-m
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Most Upvoted Answer
A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r...
Since Ish = 220/ 50 = 4.4 A
Ia = 80-4.4 = 75.6 Amp
Eb= V - IaRa
= 220- (75.6 x 0.1) = 212.44
Driving Power Eb.Ia = 212.44x75.6
= 16060 N-m
Output Power = Driving Power -  Iron and Friction Losses
= 16060-1600
= 14460 W
Armature Torque
Ta = EbIa/2πN   if N in r.p.s
Ta = (60EbIa)/(2πN)  if N in r.p.m
so; Ta = (9.55EbIa)/N   if N in r.p.s
= (9.55 x 212.44 x 75.6)/800
= 191.7 N-m
Since the whole of armature torque is not available for doing useful work, a certain part is required to overcome iron & friction losses
Hence Lost torque = 9.55x (1600/800)
=  19.1 N-m
Hence Useful Shaft torque = Ta - Tlost
= 191.7 - 19.1
= 172.6 N-m
Free Test
Community Answer
A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r...
Since Ish = 220/ 50 = 4.4 A
Ia = 80-4.4 = 75.6 Amp
Eb= V - IaRa
= 220- (75.6 x 0.1) = 212.44
Driving Power Eb.Ia = 212.44x75.6
= 16060 N-m
Output Power = Driving Power -  Iron and Friction Losses
= 16060-1600
= 14460 W
Armature Torque
Ta = EbIa/2πN   if N in r.p.s
Ta = (60EbIa)/(2πN)  if N in r.p.m
so; Ta = (9.55EbIa)/N   if N in r.p.s
= (9.55 x 212.44 x 75.6)/800
= 191.7 N-m
Since the whole of armature torque is not available for doing useful work, a certain part is required to overcome iron & friction losses
Hence Lost torque = 9.55x (1600/800)
=  19.1 N-m
Hence Useful Shaft torque = Ta - Tlost
= 191.7 - 19.1
= 172.6 N-m
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Question Description
A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r.p.m. The resistance of shunt field is 50Ω and that of armature 0.1Ω The iron and friction losses amount to 1600W. What is the shaft torque?a)152.6 N-mb)158.6 N-mc)166.4 N-md)172.6 N-mCorrect answer is option 'D'. Can you explain this answer? for SSC 2025 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r.p.m. The resistance of shunt field is 50Ω and that of armature 0.1Ω The iron and friction losses amount to 1600W. What is the shaft torque?a)152.6 N-mb)158.6 N-mc)166.4 N-md)172.6 N-mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220V d.c. shunt motor takes a total current of 80A and runs at 800 r.p.m. The resistance of shunt field is 50Ω and that of armature 0.1Ω The iron and friction losses amount to 1600W. What is the shaft torque?a)152.6 N-mb)158.6 N-mc)166.4 N-md)172.6 N-mCorrect answer is option 'D'. Can you explain this answer?.
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