2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    The emf induced in a secondary coil is 20000 V when the current breaks in the primary coil. The mutual inductance is 5 H and the current reaches to zero in\[{{10}^{-4}}s\]in the primary. The maximum current in the primary before it breaks is

    A)  0.1 A                    

    B)  0.4 A

    C)  0.6 A                                    

    D)  0.8 A

    Correct Answer: B

    Solution :

                     \[e=\frac{M{{i}_{\max }}}{t}\] Or           \[200=5\times \frac{{{i}_{\max }}}{{{10}^{-4}}}\] Or           \[{{i}_{\max }}=\frac{20000\times {{10}^{-4}}}{5}=15\times {{10}^{-4}}H\]


You need to login to perform this action.
You will be redirected in 3 sec spinner