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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

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

    A) 2A.4A                                   

    B) 4A.2A

    C) 1A.2A                                   

    D) 2 A, 3 A

    Correct Answer: A

    Solution :

    From Kirchhoff?s first law at junction P, \[{{I}_{1}}+{{I}_{2}}=6\]                                                  ...(i) From Kirchhoff?s 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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