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(D) The output voltage $V_{out}$ in a $CE$ transistor amplifier is given by the product of the output current and the load resistance.$V_{out} = I_c \

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$10\, mV$

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(D) The output voltage $V_{out}$ in a $CE$ transistor amplifier is given by the product of the output current and the load resistance.$V_{out} = I_c 	imes R_L$Given $V_{out} = 2\, V$ and $R_L = 2\, k\Omega = 2000\, \Omega$,the collector current is:$I_c = \frac{V_{out}}{R_L} = \frac{2}{2000} = 10^{-3}\, A = 1\, mA$.We know the current amplification factor $\beta = \frac{I_c}{I_b}$. Thus,the base current is:$I_b = \frac{I_c}{\beta} = \frac{1\, mA}{100} = 0.01\, mA = 10\, \mu A$.The input signal voltage $V_{in}$ is related to the base current and base resistance $R_i$ by Ohm's law:$V_{in} = I_b 	imes R_i$Given $R_i = 1\, k\Omega = 1000\, \Omega$:$V_{in} = (10 	imes 10^{-6}\, A) 	imes (1000\, \Omega) = 0.01\, V = 10\, mV$.

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 factor $(\beta)$ of the transistor is $100$,the input signal voltage is:

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