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NEET AIPMT SOLVED PAPER 1999

  • question_answer
                    The internal resistance of a cell of emf 2 V is \[0.1\,\,\Omega \]. It is connected to a resistance of \[3.9\,\,\Omega \]. The voltage across the cell will be:                                                                                                                                   

    A)                 0.5 V                     

    B)                 1.9 V     

    C)                 1.95 V

    D)                 2 V

    Correct Answer: C

    Solution :

                    Key Idea: Terminal voltage across the cell decreases due to voltage drop across internal resistance.                 The current flowing in the circuit is                 \[i=\frac{E}{R+r}\]                 Given,   E = 2 V, R = 3.9 \[\Omega \],      r = 0.1 \[\Omega \]                 \[So,i=\frac{2}{3.9+0.1}=\frac{2}{4.0}=0.5\,A\]                 The voltage drop across internal resistance,                 \[V'=ir=0.5\times 0.1\]                 1 = 0.05 V                 Thus, terminal voltage across cell is,                 \[V=E-ir=E-V'\]                 \[=2-0.05=1.95\,V\]


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