A npn transistor is connected in common emitt | vedclass

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(D) Given: Load resistance $R_L = 800 \,\,\Omega$,Input resistance $R_i = 192 \,\,\Omega$,Current amplification factor $\beta = 0.96$.The voltage gain

Vedclass · Accepted Answer

$4, 3.84$

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(D) Given: Load resistance $R_L = 800 \,\,\Omega$,Input resistance $R_i = 192 \,\,\Omega$,Current amplification factor $\beta = 0.96$.The voltage gain $(A_v)$ for a common emitter amplifier is given by the formula:$A_v = \beta 	imes \frac{R_L}{R_i}$Substituting the values:$A_v = 0.96 	imes \frac{800}{192} = 0.96 	imes 4.166... = 4$.The power gain $(A_p)$ is given by the product of current gain and voltage gain:$A_p = \beta 	imes A_v$Substituting the values:$A_p = 0.96 	imes 4 = 3.84$.Thus,the voltage gain is $4$ and the power gain is $3.84$.

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