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JEE Main & Advanced Physics Semiconducting Devices Question Bank Critical Thinking

  • question_answer
    The relation between Ip and Vp for a triode is \[{{I}_{p}}=(0.125{{V}_{p}}-7.5)\,mA\]Keeping the grid potential constant at 1V, the value of rp will be

    A)            8 kW                                        

    B)            4 kW

    C)            2 kW                                        

    D)            8 kW

    Correct Answer: D

    Solution :

                       \[{{i}_{p}}=\,[0.125\,{{V}_{p}}-7.5]\times {{10}^{-3}}\,amp\] Differentiating this equation w.r.t. V­p \[\frac{\Delta {{i}_{p}}}{\Delta {{V}_{p}}}=\,0.125\times {{10}^{-3}}\] or \[\frac{1}{{{r}_{p}}}=0.125\,\times {{10}^{-3}}\]Þ \[{{r}_{p}}\,=8\,k\Omega \]


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