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(B) In the given circuit,the base loop equation is $V_{CC} = i_B R_1 + V_{BE}$.Given $V_{CC} = 8 	ext{ V}$,$V_{BE} = 0.6 	ext{ V}$,and $i_C = 5 	ex

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$74 	ext{ k}\Omega, 1 	ext{ k}\Omega$

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(B) In the given circuit,the base loop equation is $V_{CC} = i_B R_1 + V_{BE}$.Given $V_{CC} = 8 	ext{ V}$,$V_{BE} = 0.6 	ext{ V}$,and $i_C = 5 	ext{ mA}$.The base current is $i_B = \frac{i_C}{\beta} = \frac{5 	imes 10^{-3} 	ext{ A}}{50} = 1 	imes 10^{-4} 	ext{ A}$.Substituting the values into the base loop equation:$8 = (1 	imes 10^{-4}) R_1 + 0.6$$R_1 = \frac{8 - 0.6}{1 	imes 10^{-4}} = \frac{7.4}{10^{-4}} = 74 	imes 10^3 \Omega = 74 	ext{ k}\Omega$.Now,for the collector loop,the $KVL$ equation is $V_{CC} = i_C R_2 + V_{CE}$.Substituting the values:$8 = (5 	imes 10^{-3}) R_2 + 3$$5 = (5 	imes 10^{-3}) R_2$$R_2 = \frac{5}{5 	imes 10^{-3}} = 10^3 \Omega = 1 	ext{ k}\Omega$.Thus,$R_1 = 74 	ext{ k}\Omega$ and $R_2 = 1 	ext{ k}\Omega$.

An $n-p-n$ transistor is connected in common-emitter configuration as shown in the figure. If the collector current is $5 \text{ mA}$,$V_{BE} = 0.6 \text{ V}$,$V_{CE} = 3 \text{ V}$,and the common-emitter current amplification factor $\beta = 50$,then the values of $R_1$ and $R_2$ are respectively.

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