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(C) The voltage gain $A_v$ for a common-emitter amplifier is given by $A_v = \beta 	imes \frac{R_L}{R_{in}}$.Given: $\beta = 100$, $R_L = 10 	ext{ k

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$1.0 	ext{ V}$

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(C) The voltage gain $A_v$ for a common-emitter amplifier is given by $A_v = \beta 	imes \frac{R_L}{R_{in}}$.Given: $\beta = 100$, $R_L = 10 	ext{ k}\Omega$, $R_{in} = 1 	ext{ k}\Omega$, and $V_{in} = 1 	ext{ mV}$.The output voltage $V_{out}$ is calculated as:$V_{out} = A_v 	imes V_{in} = \beta 	imes \frac{R_L}{R_{in}} 	imes V_{in}$$V_{out} = 100 	imes \frac{10 	ext{ k}\Omega}{1 	ext{ k}\Omega} 	imes 1 	ext{ mV}$$V_{out} = 100 	imes 10 	imes 1 	ext{ mV} = 1000 	ext{ mV} = 1.0 	ext{ V}$.

In the given common-emitter circuit, an $npn$ transistor with $\beta = 100$ is connected. What is the output voltage of the amplifier?

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Assertion: $A$ transistor amplifier in common emitter configuration has a low input impedance.
Reason: The base-to-emitter region is forward-biased.

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