CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    In the circuit shown, the internal resistance 3f the cell is negligible. The steady state current in the \[2\,\Omega \] resistor is:

    A)  0.6 A                                    

    B)  1.2 A

    C)  0.9 A                                    

    D)  1.5 A

    Correct Answer: C

    Solution :

    In the steady state, no current flows through the branch containing the capacitor. So, the equivalent circuit will be of the form as shown below : The effective resistance of the circuit is                 \[R=\frac{2\times 3}{2+3}+2.8\]                 \[=1.2+2.8=4\,\,\Omega \] The current through the ciruit is                 \[i=\frac{E}{R}=\frac{6}{4}\]                 \[=1.5\,A\] Let current \[{{i}_{1}}\]flows through \[2\,\Omega \]. resistance. \[\therefore \]  \[2\times {{i}_{1}}=(i-i)\times 3\] \[\Rightarrow \]               \[2{{i}_{1}}=(1.5-{{i}_{1}})\times 3\] \[\Rightarrow \]               \[2{{i}_{1}}=4.5-3{{i}_{1}}\] \[\therefore \]  \[{{i}_{1}}=0.9\,\,A\]


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