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(D) Given: Collector resistance $R_C = 2 \, k\Omega = 2000 \, \Omega$,Output voltage $V_0 = 4 \, V$,Current amplification factor $\beta = 100$,Base re

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$20$

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(D) Given: Collector resistance $R_C = 2 \, k\Omega = 2000 \, \Omega$,Output voltage $V_0 = 4 \, V$,Current amplification factor $\beta = 100$,Base resistance $R_B = 1 \, k\Omega = 1000 \, \Omega$.The voltage gain $A_v$ of a $CE$ amplifier is given by the formula: $A_v = \beta 	imes \frac{R_C}{R_B}$.Substituting the values: $A_v = 100 	imes \frac{2000}{1000} = 200$.We know that voltage gain $A_v = \frac{V_0}{V_i}$,where $V_i$ is the input signal voltage.Therefore,$V_i = \frac{V_0}{A_v} = \frac{4 \, V}{200} = 0.02 \, V$.Converting to millivolts $(mV)$: $V_i = 0.02 	imes 1000 \, mV = 20 \, mV$.

For a $CE$ transistor amplifier,the audio signal voltage across the collector resistance of $2 \, k\Omega$ is $4 \, V$. If the current amplification factor $(\beta)$ of the transistor is $100$ and the base resistance is $1 \, k\Omega$,then the input signal voltage is ....... $mV$.

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