If the value of R_e is zero then voltage gain | vedclass

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(B) The input resistance $R_b$ is given by the parallel combination of $R_1$ and $R_2$ (Thevenin equivalent resistance at the base):$R_b = \frac{R_1 R

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$2.5$

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(B) The input resistance $R_b$ is given by the parallel combination of $R_1$ and $R_2$ (Thevenin equivalent resistance at the base):$R_b = \frac{R_1 R_2}{R_1 + R_2} = \frac{5 \, k\Omega 	imes 5 \, k\Omega}{5 \, k\Omega + 5 \, k\Omega} = 2.5 \, k\Omega$.Voltage gain $A_V$ with $R_e = 0$ is given by:$A_V = \frac{\beta R_L}{R_b} = 200$.Substituting the values:$200 = \frac{\beta 	imes 5 \, k\Omega}{2.5 \, k\Omega} = 2\beta$.Therefore,$\beta = 100$.Now,for $R_e = 2 \, k\Omega$,the voltage gain $A_V'$ is given by:$A_V' = \frac{\beta R_L}{R_b + (1 + \beta) R_e}$.Substituting the values:$A_V' = \frac{100 	imes 5 \, k\Omega}{2.5 \, k\Omega + (1 + 100) 	imes 2 \, k\Omega} = \frac{500}{2.5 + 202} = \frac{500}{204.5} \approx 2.44 \approx 2.5$.

If the value of $R_e$ is zero then voltage gain is $200$. Find the voltage gain when $R_e$ is $2 \, k\Omega$.

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