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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)?

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Detailed Solution

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 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)?