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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    For the circuit shown, with \[{{R}_{1}}=1.0\Omega ,\]\[{{R}_{2}}=2.0\Omega ,{{E}_{1}}=2V\] and \[{{E}_{2}}={{E}_{3}}=4V,\]the potential difference between the points 'a' and 'b' is approximately (in V): [JEE Main 8-4-2019 Morning]

    A) 2.7                              

    B) 3.3

    C) 2.3                              

    D) 3.7

    Correct Answer: B

    Solution :

    \[{{E}_{eq}}=\frac{\frac{{{E}_{1}}}{2{{R}_{1}}}+\frac{{{E}_{2}}}{{{R}_{2}}}+\frac{{{E}_{3}}}{2{{R}_{1}}}}{\frac{1}{2{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{2{{R}_{1}}}}\]\[=\frac{\frac{2}{2}+\frac{4}{2}+\frac{4}{2}}{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}}\]                    \[=\frac{5}{\frac{3}{2}}=\frac{10}{3}=3.3\]


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