NEET Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर NEET PYQ-Current Electricity Charging Capacitors

  • question_answer
    A cell has an emf 1.5 V. When connected across an external resistance of \[2\,\,\Omega ,\] the terminal potential difference falls to 1.0 V. The internal resistance of the cell is:                                                                        [AIPMT 2000]

    A)  \[2\,\,\Omega \]            

    B)       \[1.5\,\,\Omega \]

    C)  \[1.0\,\,\Omega \]          

    D)       \[0.5\,\,\Omega \]

    Correct Answer: C

    Solution :

    [c] Internal resistance of the cell is given by
                            \[r=\left( \frac{E-V}{V} \right)R\]
                Given, \[E=1.5\text{ }V,\text{ }V=1.0\text{ }V,\text{ }R=2\,\Omega \]
                \[\therefore \]      \[r=\left( \frac{1.5-1.0}{1.0} \right)\times 2=\frac{0.5}{1.0}\times 2=1.0\,\Omega \]


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