JEE Main & Advanced Sample Paper JEE Main Sample Paper-14

  • question_answer
    The internal resistance of a ceil of emf 4 V is \[0.1\,\,\Omega \], It is connected to a resistance of 3.9 0. The voltage across the cell will be

    A)  3.9 V                    

    B)  2 V                

    C)  0.1 V                    

    D)  3.8 V

    Correct Answer: A

    Solution :

     Emf of a cell \[(E)\,=\,4\,V\] Internal resistance of a cell \[(r)=0.1\,\Omega \] External resistance \[(R)=3.9\,\Omega \] The potential drop across the cell \[V=E-I\cdot r\]                ...(i) Now, the total resistance of the circuit \[R'=r+R\] \[R'=0.1+3.9\] \[\Rightarrow \]               \[R'=4.0\,\Omega \] Hence, current in the circuit is \[I=\frac{E}{R'}\] \[\Rightarrow \]               \[I=\frac{4}{4}\,=1\,A\] Now, from Eq. (i) \[V=4\,-1\,\,\times 0.1\] \[V=4-0.1\] \[V=3.9\] volt

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