On applying a potential of $- 1\ volt$ at the grid of a triode, the following relation between plate voltage $V_{p}\ (volt)$ and plate current $I_{p}$ (in $mA$) is found $I_{p} = 0.125\ V_{p} - 7.5$. If on applying $- 3\ volt$ potential at grid and $300\ V$ potential at plate, the plate current is found to be $5mA,$ then amplification factor of the triode is

Question

Dr. Abubacker · Accepted Answer

(a) At $V_{g} = - 3V,\ V_{p} = 300\ V$ and $I_{p} = 5mA$
At $V_{g} = - 1V,$ for constant plate current
$i.e.,\ I_{p} = 5mA$
From $I_{p} = 0.125\ V_{p} - 7.5$
$\Rightarrow 5 = 0.125\ V_{p} - 7.5 \Rightarrow V_{p} = 100V$
$	herefore\ $change in plate voltage $\mathrm{\Delta}V_{p} = 300 - 100 = 200V$
Change in grid voltage $\mathrm{\Delta}V_{g} = - 1 - ( - 3) = 2V$
So, $\mu = \frac{\mathrm{\Delta}V_{p}}{\mathrm{\Delta}V_{g}} = \frac{200}{2} = 100$

Explanation

Reviewed by Dr. Abubacker

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