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(D) The power gain $(A_p)$ of a transistor amplifier is given by the product of current gain $(\beta)$ and voltage gain $(A_v)$.$A_p = \beta 	imes A_

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$250$

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(D) The power gain $(A_p)$ of a transistor amplifier is given by the product of current gain $(\beta)$ and voltage gain $(A_v)$.$A_p = \beta 	imes A_v$Given, $A_p = 5 	imes 10^6$ and $\beta = 50$.$5 	imes 10^6 = 50 	imes A_v$$A_v = \frac{5 	imes 10^6}{50} = 10^5$.Voltage gain is also defined as $A_v = \beta 	imes \frac{R_{out}}{R_{in}}$, where $R_{out}$ is the output resistance and $R_{in}$ is the input resistance.$10^5 = 50 	imes \frac{500 	imes 10^3 \Omega}{R_{in}}$$R_{in} = \frac{50 	imes 500 	imes 10^3}{10^5} = \frac{25000 	imes 10^3}{10^5} = 250 \Omega$.Thus, the input resistance is $250 \Omega$.

In a common emitter transistor amplifier, the output resistance is $500 \text{ k}\Omega$ and the current gain $\beta = 50$. If the power gain of the amplifier is $5 \times 10^6$, what is the input resistance (in $\Omega$)?

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