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(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the output circuit:$V_{CC} = I_C R_L + V_{CE}$Given $V_{CC} = 10 	ext{ V}$,$R_L = 1 	ext{ k}\Omega =

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$2 	imes 10^5 \, \Omega$

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the output circuit:$V_{CC} = I_C R_L + V_{CE}$Given $V_{CC} = 10 	ext{ V}$,$R_L = 1 	ext{ k}\Omega = 10^3 \, \Omega$,and $V_{CE} = 5 	ext{ V}$:$10 = I_C (10^3) + 5$$I_C (10^3) = 5 \implies I_C = 5 	imes 10^{-3} 	ext{ A} = 5 	ext{ mA}$Using the current gain relation $I_C = \beta I_B$:$I_B = \frac{I_C}{\beta} = \frac{5 	imes 10^{-3}}{100} = 5 	imes 10^{-5} 	ext{ A}$Applying $KVL$ to the input circuit:$V_{CC} = I_B R_B + V_{BE}$Given $V_{BE} = 0 	ext{ V}$:$10 = (5 	imes 10^{-5}) R_B + 0$$R_B = \frac{10}{5 	imes 10^{-5}} = 2 	imes 10^5 \, \Omega$

In the circuit shown in the figure,the current gain $\beta = 100$ for an $npn$ transistor. What should be the base resistor $R_B$ so that $V_{CE} = 5 \text{ V}$,given $V_{BE} = 0 \text{ V}$?

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