In an N-P-N transistor circuit,the common emi | vedclass

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(A) Given $I_C = 4mA = 4 	imes 10^{-3} A$. Applying Kirchhoff's voltage law to the collector circuit:$V_{CC} = I_C R_L + V_{CE}$$8V = (4 	imes 10^{-

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$1 k\Omega, 185 k\Omega$

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(A) Given $I_C = 4mA = 4 	imes 10^{-3} A$. Applying Kirchhoff's voltage law to the collector circuit:$V_{CC} = I_C R_L + V_{CE}$$8V = (4 	imes 10^{-3} A) R_L + 4V$$4V = (4 	imes 10^{-3} A) R_L$$R_L = \frac{4}{4 	imes 10^{-3}} \Omega = 1000 \Omega = 1 k\Omega$Now,for the base circuit,the base current $I_B$ is given by:$I_B = \frac{I_C}{\beta} = \frac{4mA}{100} = 0.04mA = 4 	imes 10^{-5} A$Applying Kirchhoff's voltage law to the base circuit:$V_{CC} = I_B R_B + V_{BE}$$8V = (4 	imes 10^{-5} A) R_B + 0.6V$$7.4V = (4 	imes 10^{-5} A) R_B$$R_B = \frac{7.4}{4 	imes 10^{-5}} \Omega = 1.85 	imes 10^5 \Omega = 185 k\Omega$Thus,$R_L = 1 k\Omega$ and $R_B = 185 k\Omega$.

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