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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2008

  • question_answer
        A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistances\[2\,\Omega \]each and leaves by the corner R. Then the current\[{{I}_{1}}\]and\[{{I}_{2}}\]are

    A)  2 A, 4 A                               

    B)  4 A, 2 A

    C)  1A, 2A                                 

    D)  2A, 3A

    Correct Answer: A

    Solution :

                    From Kirchhoffs first law at junction P, \[{{I}_{1}}+{{I}_{2}}=6\]                            ...(i) From Kirchhoffs second law to the closed circuit PQRP,                 \[-2{{I}_{1}}-2{{I}_{1}}+2{{I}_{2}}=0\] \[\Rightarrow \]               \[-4{{I}_{1}}+2{{I}_{2}}=0\] \[\Rightarrow \]               \[2{{I}_{1}}-{{I}_{2}}=0\]                                               ?.(ii) Adding Eqs. (i) and (ii), we get                 \[3{{I}_{1}}=6\] \[\Rightarrow \]               \[{{I}_{1}}=2A\] From Eq. (i), \[{{I}_{2}}=6-2=4A\]


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