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(A) Given: Current gain $\alpha = 0.98$, Load resistance $R_L = 5 	ext{ k}\Omega = 5000 \Omega$, Input resistance $R_{in} = 70 \Omega$.The voltage ga

Vedclass · Accepted Answer

$70, 68.6$

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(A) Given: Current gain $\alpha = 0.98$, Load resistance $R_L = 5 	ext{ k}\Omega = 5000 \Omega$, Input resistance $R_{in} = 70 \Omega$.The voltage gain $A_v$ is given by the formula: $A_v = \alpha 	imes \frac{R_L}{R_{in}}$.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 and voltage gain: $A_p = \alpha 	imes A_v$.Substituting the values: $A_p = 0.98 	imes 70 = 68.6$.Therefore, the voltage gain is $70$ and the power gain is $68.6$.

$A$ transistor is used as an amplifier in a common-base configuration. The load resistance is $5 \text{ k}\Omega$, the current gain is $0.98$, and the input resistance is $70 \Omega$. The voltage gain and power gain are, respectively:

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