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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    A generator has armature resistance of \[0.1\Omega \]and develops an induced emf of 120 F when driven at its rated speed. Its terminal voltage when a current of 50 A is being drawn is     JEE Main  Online Paper (Held On 19  May  2012)

    A) 120 V                                    

    B)                        5V

    C)                        115 V       

    D)                        70 V

    Correct Answer: C

    Solution :

                    Armature resistance \[R=0.1\Omega \] Induced emf, e = 120V Current drown, I = 50 A We know that, e = V + IR or, V= e - IR = 120 - 50 \[\times \] 0.1 = 115V                                


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