The input resistance of a common-emitter ampl | vedclass

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(A) Given:Input resistance $R_i = 2 \, k\Omega = 2 	imes 10^3 \, \Omega$.$A$.$C$. current gain $\beta = 20$.Load resistance $R_L = 5 \, k\Omega$.The

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$0.01$

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(A) Given:Input resistance $R_i = 2 \, k\Omega = 2 	imes 10^3 \, \Omega$.$A$.$C$. current gain $\beta = 20$.Load resistance $R_L = 5 \, k\Omega$.The transconductance $g_m$ of a transistor is defined as the ratio of the output current change to the input voltage change,which is given by the formula:$g_m = \frac{\beta}{R_i}$Substituting the given values:$g_m = \frac{20}{2 	imes 10^3 \, \Omega}$$g_m = 10 	imes 10^{-3} \, \Omega^{-1}$$g_m = 0.01 \, \Omega^{-1}$Thus,the transconductance is $0.01 \, \Omega^{-1}$.

The input resistance of a common-emitter amplifier is $2 \, k\Omega$ and a.c. current gain is $20$. If the load resistor used is $5 \, k\Omega$,calculate the transconductance of the transistor used in $\Omega^{-1}$.

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