JEE Main & Advanced Sample Paper JEE Main Sample Paper-34

  • question_answer
    Given\[{{R}_{1}}=1\Omega \,\,=2\Omega ,\,\,\,\,{{C}_{1}}=2\mu F,\] \[{{C}_{2}}=4\mu F\], the time constant (in \[\mu s\]) for the circuit I, II, III are respectively.

    A)  \[18,\frac{8}{9},4\]                        

    B)  \[18,4\frac{8}{9}\]

    C)  \[4,\frac{8}{9},18\]                        

    D)  \[\frac{8}{9},18,4\]

    Correct Answer: D

    Solution :

    since \[\tau =RC\] where R is effective resistance across the effective capacitor while the battery or batteries are shortcircuited and C is effective capacitance.             In 1, \[R=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{2}{3}\Omega \]  and \[C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{4}{3}\,\mu F\] \[\therefore \,\,\tau \,\,=RC\,=\frac{8}{9}\,\mu s\] In II, \[R={{R}_{1}}+{{R}_{2}}=3\Omega \] and \[C={{C}_{1}}+{{C}_{2}}\,=6\mu F\] \[\therefore \,\,\tau \,=RC=18\,\mu s\] \[R=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\,=\frac{2}{3}\,\Omega \] and \[C={{C}_{1}}+{{C}_{2}}=6\mu F\]      \[\therefore \,\,\tau =RC\,=4\mu s\]

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