In a transistor amplifier, a change of 0.2 te | vedclass

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(B) The current gain $\beta$ is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.$\

Vedclass · Accepted Answer

$6 	ext{ k}\Omega$

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(B) The current gain $\beta$ is defined as the ratio of the change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$.$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 	ext{ mA}}{0.2 	ext{ mA}} = 25$.The voltage gain $A_V$ is given by the formula $A_V = \beta 	imes \frac{R_L}{R_{in}}$, where $R_L$ is the load resistance and $R_{in}$ is the input resistance.Given $A_V = 75$, $\beta = 25$, and $R_{in} = 2 	ext{ k}\Omega$.Substituting the values: $75 = 25 	imes \frac{R_L}{2 	ext{ k}\Omega}$.$3 = \frac{R_L}{2 	ext{ k}\Omega}$.$R_L = 3 	imes 2 	ext{ k}\Omega = 6 	ext{ k}\Omega$.Therefore, the load resistance is $6 	ext{ k}\Omega$.

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

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