In the shown common emitter amplifier circuit | vedclass

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$8.7 - I_B(200 	imes 10^3) - V_{BE} = 0$$8.7 - I_B(200 	imes 10^3) - 0.7 = 0$$8 = I_B

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$2.5\,k\Omega$

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$8.7 - I_B(200 	imes 10^3) - V_{BE} = 0$$8.7 - I_B(200 	imes 10^3) - 0.7 = 0$$8 = I_B(200 	imes 10^3)$$I_B = \frac{8}{200 	imes 10^3} = 4 	imes 10^{-5}\,A = 40\,\mu A$Given $\beta = 80$,we find the collector current $I_C$:$I_C = \beta I_B = 80 	imes 4 	imes 10^{-5} = 320 	imes 10^{-5} = 3.2 	imes 10^{-3}\,A = 3.2\,mA$Applying $KVL$ to the output loop:$11 - I_C R_C - V_{CE} = 0$$11 - (3.2 	imes 10^{-3}) R_C - 3 = 0$$8 = (3.2 	imes 10^{-3}) R_C$$R_C = \frac{8}{3.2 	imes 10^{-3}} = 2.5 	imes 10^3\,\Omega = 2.5\,k\Omega$

In the shown common emitter amplifier circuit,$\beta = 80$ and $V_{BE} = 0.7\,V$. The value of $R_C$ is:

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