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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    An uniformly wound solenoid coil of self-inductance \[1.8\times {{10}^{-4}}H\] and resistance \[6\Omega \] is broken up into two identical coils. These identical coils are then connected in parallel across a \[12V\]battery of negligible resistance. The time constant for the current in the circuit is :

    A)  \[0.1\times {{10}^{-4}}s\]                   

    B)  \[0.2\times {{10}^{-4}}s\]

    C)  \[0.3\times {{10}^{-4}}s\]                   

    D)  \[0.4\times {{10}^{-4}}s\]

    Correct Answer: C

    Solution :

    Given, \[L=1.8\,\times {{10}^{-4}}\,H,\,\,R=6\Omega \] Self-inductance of each coil \[=\frac{1.8\,\times {{10}^{-4}}}{2}H\] Resistance of each coil \[=\frac{6\Omega }{2}=3\Omega \] Coils are then connected in parallel \[\therefore \,\,L'=\frac{\frac{1.8}{2}\times {{10}^{-4}}\,\times \frac{1.8}{2}\times {{10}^{-4}}}{\frac{1.8}{2}\times {{10}^{-4}}+\,\frac{1.8}{2}\,\times {{10}^{-4}}}\] \[=0.45\,\times \,{{10}^{-4}}H\]and \[R'\,=\frac{3\times 3}{3+3}\,=1.5\Omega \]time constant \[\frac{L'}{R'}\,=\frac{0.45\,\times {{10}^{-4}}}{1.5}\] \[=0.3\,\times {{10}^{-4}}s\]


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