JEE Main & Advanced Physics Electrostatics & Capacitance NEET PYQ - Electrostatics and Capacitance

  • question_answer
    A student measures the terminal potential difference (V) of a cell (of emf s and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, then respectively, equal      [AIPMT (S) 2009]

    A) \[\varepsilon \] and \[-r\]

    B) \[-r\] and \[\varepsilon \]

    C) \[r\] and \[-\,\varepsilon \]

    D) \[-\,\varepsilon \] and \[r\]

    Correct Answer: B

    Solution :

    [b] According to Ohm's law \[\frac{dV}{dI}=-r\] and \[V=\varepsilon \] if \[I=0\]            [As \[V+Ir=\varepsilon \]]
    \[\therefore \] Slope of the graph \[=-r\] and intercept \[=\varepsilon \]


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