Question
For a given triode $\mu = 20$. The load resistance is $1.5$ times the anode resistance. The maximum gain will be
✓
Answer
(b)Voltage gain ${A_v} = \frac{\mu }{{1 + \frac{{{r_p}}}{{{R_L}}}}}$
$ \Rightarrow {A_v} = \frac{\mu }{{1 + \frac{{{r_p}}}{{1.5\,{r_p}}}}} = \frac{3}{5}\mu $$ = \frac{3}{5} \times 20 = 12$.
$ \Rightarrow {A_v} = \frac{\mu }{{1 + \frac{{{r_p}}}{{1.5\,{r_p}}}}} = \frac{3}{5}\mu $$ = \frac{3}{5} \times 20 = 12$.
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