A simple electric motor has an armature resistance of $$1$$Ω and runs from a dc source of $$12$$ volt. When running unloaded it draws a current of $$2A$$. When a certain load is connected, its speed becomes $$one-half$$ of its unloaded value. What is the new value of current drawn (in$$ $$ampere)?

Question

A simple electric motor has an armature resistance of $$1$$Ω and runs from a dc source of $$12$$ volt. When running unloaded it draws a current of $$2A$$. When a certain load is connected, its speed becomes $$one-half$$ of its unloaded value. What is the new value of current drawn (in$$ $$ampere)?

Infinity Learn · Accepted Answer

Let initial e.m.f. induced $$=e$$∴ Initial current $$i=E$$−eR ⇒ $$2=12$$−$$e1$$⇒$$e=12$$−$$2=10$$ V As e ∝ω when speed is halved, the value of induced e.m.f. becomes ⇒$$e2=102=5$$ V∴ New value of current $$=i$$'$$=E$$−$$eR=12$$−$$51=7$$ A

Magnetic flux and induced emf

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A simple electric motor has an armature resistance of $1 Ω$ and runs from a dc source of 12 volt. When running unloaded it draws a current of 2A. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn (in ampere)?

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Magnetic flux and induced emf

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A simple electric motor has an armature resistance of 1Ω and runs from a dc source of 12 volt. When running unloaded it draws a current of 2A. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn (in ampere)?

Let initial e.m.f. induced =e∴ Initial current i=E−eR ⇒ 2=12−e1⇒e=12−2=10 V As e ∝ω when speed is halved, the value of induced e.m.f. becomes ⇒e2=102=5 V∴ New value of current =i'=E−eR=12−51=7 A

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A simple electric motor has an armature resistance of $1 Ω$ and runs from a dc source of 12 volt. When running unloaded it draws a current of 2A. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn (in ampere)?