In a transistor in CE configuration,the ratio | vedclass

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(D) For a transistor in $CE$ configuration,the voltage gain $(A_v)$ is given by $A_v = \beta 	imes \frac{R_L}{R_i}$,where $\beta$ is the current gain

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$\beta$

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(D) For a transistor in $CE$ configuration,the voltage gain $(A_v)$ is given by $A_v = \beta 	imes \frac{R_L}{R_i}$,where $\beta$ is the current gain,$R_L$ is the load resistance,and $R_i$ is the input resistance.The power gain $(A_p)$ is given by $A_p = \beta^2 	imes \frac{R_L}{R_i}$.Therefore,the ratio of power gain to voltage gain is:$\frac{A_p}{A_v} = \frac{\beta^2 	imes (R_L / R_i)}{\beta 	imes (R_L / R_i)} = \beta$.Thus,the correct option is $D$.

In a transistor in $CE$ configuration,the ratio of power gain to voltage gain is

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