JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Grouping of Resistances

  • question_answer
    Two wires of same metal have the same length but their cross-sections are in the ratio \[3:1\]. They are joined in series. The resistance of the thicker wire is \[10\,\Omega \]. The total resistance of the combination will be        [CBSE PMT 1995]

    A)                    \[40\,\Omega \]

    B)                                      \[\frac{40}{3}\Omega \]

    C)                    \[\frac{5}{2}\Omega \]    

    D)            \[100\,\Omega \]

    Correct Answer: A

    Solution :

                 For same material and same length \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{3}{2}\] Þ \[{{R}_{2}}=3{{R}_{1}}\] Resistance of thick wire \[{{R}_{1}}=10\Omega \] \ Resistance of thin wire \[{{R}_{2}}=30\Omega \] Total resistance in series = 10 + 30 = 40 W


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