JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Different Measuring Instruments

  • question_answer
    We have a galvanometer of resistance \[25\,\Omega \]. It is shunted by a \[2.5\,\Omega \] wire. The part of total current that flows through the galvanometer is given as                                               [AFMC 1998; MH CET 1999; Pb. PMT 2002]

    A)                    \[\frac{I}{{{I}_{0}}}=\frac{1}{11}\]

    B)                                      \[\frac{I}{{{I}_{0}}}=\frac{1}{10}\]

    C)                    \[\frac{I}{{{I}_{0}}}=\frac{3}{11}\]                           

    D)            \[\frac{I}{{{I}_{0}}}=\frac{4}{11}\]

    Correct Answer: A

    Solution :

                 \[\frac{i}{{{i}_{g}}}=\frac{G+S}{S}\]Þ \[\frac{{{i}_{g}}}{i}=\frac{S}{G+S}=\frac{2.5}{27.5}=\frac{1}{11}\]


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