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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Self Evaluation Test - Electromagnetic Induction

  • question_answer
    A simple electric motor has an armature resistance of \[1\Omega ~\] and runs from a dc source of 12 volt. When running unloaded it draws a current of 2 amp. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn?

    A) 7 A      

    B) 3 A   

    C) 5 A      

    D) 4 A

    Correct Answer: A

    Solution :

    [a] Let initial e.m.f. induced =e \[\therefore \]  Initial current \[i=\frac{E-e}{R}\]  i.e., \[2=\frac{12-e}{1}\] This gives \[e=12-2\text{=}10\] volt. As \[e\propto \omega \]. when speed is halved, the value of induced e.m.f. becomes \[\frac{e}{2}=\frac{10}{2}=5\,volt\] \[\therefore \]   New value of current \[i'=\frac{E-e}{R}=\frac{12-5}{1}=7A\]


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