BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    In the given figure the steady state current in the circuit is

    A) Zero                                     

    B) 0.6 A                    

    C) 0.9 A                    

    D) 1.5 A

    Correct Answer: D

    Solution :

    Key Idea: In the steady state there is no current in the capacitor branch. In the given circuit the resistors of \[2\,\Omega \] and \[3\,\Omega \] are connected in parallel hence, equivalent resistance is \[\frac{1}{R}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\] \[\therefore \]                  \[R=\frac{6}{5}\Omega \] Also in steady state, the circuit is shown as. Resistors of \[\frac{6}{5}\Omega \] and \[2.8\,\Omega \] are connected in series Hence,                  \[R\,=\frac{6}{5}\Omega +2.8\,\Omega \]                                 \[=1.2\Omega +2.8\Omega \]                                 \[=4.0\,\Omega \] From Ohms law, Current \[i=\frac{V}{R}=\frac{6}{4.0}=1.5A\]


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