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(A) The output $ac$ voltage is $V_{out} = 2.0\,V$. The collector resistance is $R_C = 2.0\,k\Omega = 2000\,\Omega$.The $ac$ collector current is $i_C

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$14$

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(A) The output $ac$ voltage is $V_{out} = 2.0\,V$. The collector resistance is $R_C = 2.0\,k\Omega = 2000\,\Omega$.The $ac$ collector current is $i_C = V_{out} / R_C = 2.0\,V / 2000\,\Omega = 1.0\,mA$.The signal base current is $i_B = i_C / \beta = 1.0\,mA / 100 = 0.010\,mA$.The $dc$ base current $I_B$ is given as $10$ times the signal current: $I_B = 10 	imes 0.010\,mA = 0.10\,mA$.Using the base circuit equation $V_{BB} = I_B R_B + V_{BE}$,we assume the standard base-emitter voltage $V_{BE} = 0.6\,V$.$R_B = (V_{BB} - V_{BE}) / I_B = (2.0\,V - 0.6\,V) / 0.10\,mA = 1.4\,V / 0.10\,mA = 14\,k\Omega$.

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