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(D) For a $CE$ amplifier,the voltage gain $A_V$ is given by $A_V = g_m R_L$,where $g_m$ is the transconductance and $R_L$ is the load resistance.Given

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$2/3 G$

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(D) For a $CE$ amplifier,the voltage gain $A_V$ is given by $A_V = g_m R_L$,where $g_m$ is the transconductance and $R_L$ is the load resistance.Given for the first transistor: $g_{m1} = 0.03 \ mho$ and $\beta_1 = 25$. The voltage gain is $G = g_{m1} R_L$.Given for the second transistor: $g_{m2} = 0.02 \ mho$ and $\beta_2 = 20$. The new voltage gain is $G' = g_{m2} R_L$.Taking the ratio of the two gains: $\frac{G'}{G} = \frac{g_{m2} R_L}{g_{m1} R_L} = \frac{g_{m2}}{g_{m1}}$.Substituting the values: $\frac{G'}{G} = \frac{0.02}{0.03} = \frac{2}{3}$.Therefore,the new voltage gain is $G' = \frac{2}{3} G$.

$A$ $CE$ amplifier has a voltage gain of $G$. The transistor used has a transconductance of $0.03 \ mho$ and a current gain of $25$. If this transistor is replaced by another transistor with a transconductance of $0.02 \ mho$ and a current gain of $20$,the new voltage gain will be ........

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