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JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    An AC source producing emf \[e={{e}_{0}}[\cos (100\pi {{s}^{-1}})t+\cos (500\pi {{s}^{-1}})t]\] is connected in series with a capacitor and resistor. The steady-state current in the circuit is found to be \[i={{i}_{1}}\cos [(100\pi {{s}^{-1}})t+{{\phi }_{1}}]+{{i}_{2}}\cos \]\[[(500\pi {{s}^{-1}})t+{{\phi }_{2}}]\]

    A)  \[{{i}_{1}}>{{i}_{2}}\]                    

    B)  \[{{i}_{1}}={{i}_{2}}\]

    C)  \[{{i}_{1}}<{{i}_{2}}\]

    D)  The information is insufficient to find the relation between \[{{i}_{1}}\] and \[{{i}_{2}}\]

    Correct Answer: C

    Solution :

    \[{{i}_{1}}=\frac{{{E}_{0}}}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{{{\omega }_{1}}C} \right)}^{2}}}}=\frac{{{E}_{0}}}{{{Z}_{1}}},\]where \[{{\omega }_{1}}=100\pi \] \[{{i}_{2}}=\frac{{{E}_{0}}}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{{{\omega }_{1}}C} \right)}^{2}}}}=\frac{{{E}_{0}}}{{{Z}_{2}}},\] where \[{{\omega }_{2}}=500\pi \] So,    \[{{Z}_{1}}>{{Z}_{2}},\]so\[{{i}_{1}}<{{i}_{2}}\]


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