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(D) The current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:$\beta = \frac{\Delta I_C}{\Delt

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$6 	ext{ k}\Omega$

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(D) The current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:$\beta = \frac{\Delta I_C}{\Delta I_b} = \frac{5 	ext{ mA}}{0.2 	ext{ mA}} = 25$The voltage gain $A_v$ for a common emitter amplifier is given by the formula:$A_v = \beta 	imes \frac{R_L}{R_i}$Given that $A_v = 75$,$R_i = 2 	ext{ k}\Omega$,and $\beta = 25$,we can solve for the load resistance $R_L$:$75 = 25 	imes \frac{R_L}{2 	ext{ k}\Omega}$Rearranging the equation to solve for $R_L$:$R_L = \frac{75 	imes 2 	ext{ k}\Omega}{25}$$R_L = 3 	imes 2 	ext{ k}\Omega = 6 	ext{ k}\Omega$Therefore,the load resistance is $6 	ext{ k}\Omega$.

In a common emitter amplifier,a change of $0.2 \text{ mA}$ in the base current causes a change of $5 \text{ mA}$ in the collector current. If the input resistance is $2 \text{ k}\Omega$ and the voltage gain is $75$,what is the load resistance used in the circuit?

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