On applying a potential of -1 V at the grid | vedclass

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(A) Given,for $V_g = -1 \ V$,the relation is $I_p = 0.125 V_p - 7.5$.For $V_g = -3 \ V$ and $V_p = 300 \ V$,the plate current $I_p = 5 \ mA$.To find t

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$100$

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(A) Given,for $V_g = -1 \ V$,the relation is $I_p = 0.125 V_p - 7.5$.For $V_g = -3 \ V$ and $V_p = 300 \ V$,the plate current $I_p = 5 \ mA$.To find the amplification factor $\mu = \left( \frac{\Delta V_p}{\Delta V_g} ight)_{I_p = 	ext{const}}$,we first find the plate voltage $V_p'$ required to maintain $I_p = 5 \ mA$ at $V_g = -1 \ V$.Substituting $I_p = 5 \ mA$ in the given equation:$5 = 0.125 V_p' - 7.5$$12.5 = 0.125 V_p'$$V_p' = \frac{12.5}{0.125} = 100 \ V$.Now,the change in plate voltage $\Delta V_p = 300 \ V - 100 \ V = 200 \ V$.The change in grid voltage $\Delta V_g = -1 \ V - (-3 \ V) = 2 \ V$.Therefore,the amplification factor $\mu = \frac{\Delta V_p}{\Delta V_g} = \frac{200 \ V}{2 \ V} = 100$.

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