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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • 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)  2 A, 4 A               

    B)  4 A, 2 A

    C)  1 A, 2 A                               

    D)  2 A, 3 A

    Correct Answer: A

    Solution :

    From Kirchhoffs first law at jucntion 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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