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(C) The voltage gain $(A_v)$ of a common-emitter amplifier is given by the formula: $A_v = \beta 	imes \frac{R_{	ext{out}}}{R_{	ext{in}}}$.First,we

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$990$

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(C) The voltage gain $(A_v)$ of a common-emitter amplifier is given by the formula: $A_v = \beta 	imes \frac{R_{	ext{out}}}{R_{	ext{in}}}$.First,we calculate the current gain $\beta$ using the given value of $\alpha = 0.99$:$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99$.Now,substitute the values of $\beta$,$R_{	ext{out}} = 10 \, k\Omega$,and $R_{	ext{in}} = 1 \, k\Omega$ into the voltage gain formula:$A_v = 99 	imes \frac{10 \, k\Omega}{1 \, k\Omega} = 99 	imes 10 = 990$.Therefore,the voltage gain is $990$.

$A$ common-emitter amplifier is designed with a $n-p-n$ transistor $(\alpha = 0.99)$. The input impedance is $1 \, k\Omega$ and the load resistance is $10 \, k\Omega$. The voltage gain will be

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