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(D) The voltage gain $A_v$ of a $CE$ transistor amplifier is given by the product of current gain $\beta$ and the resistance gain $\frac{R_L}{R_i}$.$A

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$10\,mV$

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(D) The voltage gain $A_v$ of a $CE$ transistor amplifier is given by the product of current gain $\beta$ and the resistance gain $\frac{R_L}{R_i}$.$A_v = \beta 	imes \frac{R_L}{R_i}$Given: $\beta = 100$,$R_L = 2\,k\Omega$,$R_i = 1\,k\Omega$,and $V_{out} = 2\,V$.The relationship between input voltage $V_{in}$ and output voltage $V_{out}$ is $V_{out} = A_v 	imes V_{in}$.Substituting the values: $2 = (100 	imes \frac{2\,k\Omega}{1\,k\Omega}) 	imes V_{in}$.$2 = 100 	imes 2 	imes V_{in}$.$2 = 200 	imes V_{in}$.$V_{in} = \frac{2}{200} = 0.01\,V$.Converting to millivolts: $V_{in} = 0.01 	imes 1000\,mV = 10\,mV$.

In a $CE$ transistor amplifier,the audio signal voltage across the collector resistance of $2\,k\Omega$ is $2\,V$. If the base resistance is $1\,k\Omega$ and the current amplification of the transistor is $100$,the input signal voltage is

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