JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Self Evaluation Test - Current Electricity

  • question_answer
    Two sources of equal emf are connected to an external resistance R. The identical resistance of the two sources are \[{{R}_{1}}\] and \[{{R}_{2}}\]\[\left( {{R}_{2}}>{{R}_{1}} \right)\]. If the potential difference across the source having internal resistance \[{{R}_{2}}\]is zero, then

    A) \[R={{R}_{2}}-{{R}_{1}}\]

    B) \[R={{R}_{2}}\times \left( {{R}_{1}}+{{R}_{2}} \right)/\left( {{R}_{2}}-{{R}_{1}} \right)\]

    C) \[R={{R}_{1}}{{R}_{2}}/\left( {{R}_{2}}-{{R}_{1}} \right)\]

    D) \[R={{R}_{1}}{{R}_{2}}/\left( {{R}_{1}}-{{R}_{2}} \right)\]

    Correct Answer: C

    Solution :

    [c] \[I=\frac{2\varepsilon }{R+{{R}_{1}}+{{R}_{2}}}\] Pot. difference across second cell \[=V=\varepsilon -I{{R}_{2}}=0\] \[\varepsilon =\frac{2\varepsilon }{R+{{R}_{1}}+{{R}_{2}}}.{{R}_{2}}=0\] \[R+{{R}_{1}}+{{R}_{2}}-2{{R}_{2}}=0\] \[R+{{R}_{1}}-{{R}_{2}}=0\text{       }\therefore R={{R}_{2}}-{{R}_{1}}\]


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