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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    Switch S of the circuit shown in figure is closed at\[t=0\]. If \[\varepsilon \] denotes the induced emf in L and I the current flowing through the circuit at time t, then which of the following graphs is correct:             

    A)                         

    B)  

    C)                        

    D)  

    Correct Answer: C

    Solution :

    \[\varepsilon E-IR=E-E(1-{{e}^{-1/t}})=E{{e}^{-1/t}}\] Therefore \[\varepsilon \] vs \[t\]graph is represented by option [c]


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