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JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Kirchhoff's Law, Cells

  • question_answer
    A cell whose e.m.f. is 2 V and internal resistance is \[0.1\,\Omega \], is connected with a resistance of\[3.9\,\Omega \]. The voltage across the cell terminal will be [CPMT 1990; MP PET 1993; CBSE PMT 1999; AFMC 1999; Pb. PMT 2000; AIIMS 2001]

    A)                    \[0.50\,V\]

    B)                                      \[1.90\,V\]

    C)                    \[1.95\,V\]                    

    D)            \[2.00\,V\]

    Correct Answer: C

    Solution :

               The voltage across cell terminal will be given by \[=\frac{E}{R+r}\times R\] \[=\frac{2}{(3.9+0.1)}\times 3.9=1.95V\]


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