JEE Main & Advanced Sample Paper JEE Main - Mock Test - 13

  • question_answer
    A current of 4A flows in a coil when connected to a \[12\,V\] dc source. If the same coil is connected to a \[12\,V\], \[50\text{ }rad/s\text{ }a.c.\]source, a current of \[2.4A\] flows in the circuit. Determine the inductance of the coil.

    A) \[0.08\text{ }H\]            

    B) \[0.04\text{ }H\]   

    C) \[0.02\text{ }H\]                        

    D) \[1H\]

    Correct Answer: A

    Solution :

    A coil consists of an inductance (L) and a resistance (R).
    In dc only resistance is effective. Hence, \[R=\frac{V}{i}=\frac{12}{4}=3\Omega \]
    In ac, \[{{i}_{rms}}=\frac{{{V}_{rms}}}{Z}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\]
    \[\therefore \,\,{{L}^{2}}=\frac{1}{{{\omega }^{2}}}\left[ {{\left( \frac{{{V}_{rms}}}{{{i}_{rms}}} \right)}^{2}}-{{R}^{2}} \right]\]
    \[\Rightarrow \,\,\,\,L=\frac{1}{\omega }\,\sqrt{{{\left( \frac{{{V}_{rms}}}{{{i}_{rms}}} \right)}^{2}}-{{R}^{2}}}=\frac{1}{50}\sqrt{{{\left( \frac{12}{2.4} \right)}^{2}}-{{(3)}^{2}}}\]
    \[=0.08\] henry


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