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(D) Voltage gain is given by the product of current gain and resistance gain.$A_{v} = \beta 	imes \frac{R_{	ext{out}}}{R_{	ext{in}}}$Since transcon

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$\frac{2}{3} G$

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(D) Voltage gain is given by the product of current gain and resistance gain.$A_{v} = \beta 	imes \frac{R_{	ext{out}}}{R_{	ext{in}}}$Since transconductance $g_{m} = \frac{\beta}{R_{	ext{in}}}$,we can write $R_{	ext{in}} = \frac{\beta}{g_{m}}$.Substituting this into the voltage gain formula: $A_{v} = \beta 	imes \frac{R_{	ext{out}}}{\beta / g_{m}} = g_{m} R_{	ext{out}}$.For the first transistor: $G = 0.03 	imes R_{	ext{out}}$ (Equation $i$).For the second transistor: $G' = 0.02 	imes R_{	ext{out}}$ (Equation $ii$).Dividing Equation $ii$ by Equation $i$: $\frac{G'}{G} = \frac{0.02}{0.03} = \frac{2}{3}$.Therefore,the new voltage gain is $G' = \frac{2}{3} G$.

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

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