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(C) Given: Load resistance $R_L = 3 k\Omega = 3000 \Omega$. Input voltage $V_i = 30 mV = 30 	imes 10^{-3} V$. Change in base current $\Delta I_B = 30

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$30000$

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(C) Given: Load resistance $R_L = 3 k\Omega = 3000 \Omega$. Input voltage $V_i = 30 mV = 30 	imes 10^{-3} V$. Change in base current $\Delta I_B = 30 \mu A = 30 	imes 10^{-6} A = 3 	imes 10^{-5} A$. Change in collector current $\Delta I_C = 3 mA = 3 	imes 10^{-3} A$. First, calculate the current gain $\beta$: $\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3 	imes 10^{-3}}{3 	imes 10^{-5}} = 100$. Next, calculate the input resistance $R_{in}$: $R_{in} = \frac{V_i}{\Delta I_B} = \frac{30 	imes 10^{-3}}{30 	imes 10^{-6}} = 1000 \Omega$. The power gain $A_P$ is given by the formula: $A_P = \beta^2 	imes \frac{R_L}{R_{in}}$. Substituting the values: $A_P = (100)^2 	imes \frac{3000}{1000} = 10000 	imes 3 = 30000$. Thus, the power gain is $30000$.

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