A transistor is used as an amplifier in CB mo | vedclass

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$(A)$ The voltage gain $(A_V)$ for a transistor in $CB$ mode is given by the formula: $A_V = \alpha 	imes \frac{R_L}{R_i}$.Given: $\alpha = 0.98$, $R

Vedclass · Accepted Answer

$70, 68.6$

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$(A)$ The voltage gain $(A_V)$ for a transistor in $CB$ mode is given by the formula: $A_V = \alpha 	imes \frac{R_L}{R_i}$.Given: $\alpha = 0.98$, $R_L = 5 \, k\Omega = 5000 \, \Omega$, and $R_i = 70 \, \Omega$.Substituting the values: $A_V = 0.98 	imes \frac{5000}{70} = 0.98 	imes 71.428 \approx 70$.The power gain $(A_p)$ is given by the product of current gain $(\alpha)$ and voltage gain $(A_V)$: $A_p = \alpha 	imes A_V$.Substituting the values: $A_p = 0.98 	imes 70 = 68.6$.Thus, the voltage gain is $70$ and the power gain is $68.6$.

$A$ transistor is used as an amplifier in $CB$ mode with a load resistance of $5 \, k\Omega$. The current gain of the amplifier is $0.98$ and the input resistance is $70 \, \Omega$. The voltage gain and power gain respectively are:

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