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(B) The voltage gain $A_{V}$ of a common emitter amplifier is given by the formula $A_{V} = g_{m} R_{L}$,where $g_{m}$ is the transconductance and $R_

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$\frac{2}{3}\,G$

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(B) The voltage gain $A_{V}$ of a common emitter amplifier is given by the formula $A_{V} = g_{m} R_{L}$,where $g_{m}$ is the transconductance and $R_{L}$ is the load resistance.For the first transistor: $A_{V1} = G = g_{m1} R_{L} = 0.03 	imes R_{L}$.For the second transistor: $A_{V2} = g_{m2} R_{L} = 0.02 	imes R_{L}$.Taking the ratio of the two gains:$\frac{A_{V2}}{A_{V1}} = \frac{g_{m2} R_{L}}{g_{m1} R_{L}} = \frac{g_{m2}}{g_{m1}}$.Substituting the given values:$\frac{A_{V2}}{G} = \frac{0.02}{0.03} = \frac{2}{3}$.Therefore,the new voltage gain is $A_{V2} = \frac{2}{3} G$.

In a common emitter $(CE)$ amplifier having a voltage gain $G$,the transistor used has a transconductance of $0.03\,mho$ and a current gain of $25$. If the above transistor is replaced with another one with a transconductance of $0.02\,mho$ and a current gain of $20$,the new voltage gain will be:

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