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(B) From the given circuit,we apply Kirchhoff's Voltage Law $(KVL)$ to the input loop (base-emitter loop).The loop equation is: $V_{CC} - I_B R_B - V_

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$0.2$

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(B) From the given circuit,we apply Kirchhoff's Voltage Law $(KVL)$ to the input loop (base-emitter loop).The loop equation is: $V_{CC} - I_B R_B - V_{BE} = 0$.Given values are $V_{CC} = 10 \, V$,$R_B = 245 \, k\Omega = 245 	imes 10^3 \, \Omega$,and $I_B = 40 \, \mu A = 40 	imes 10^{-6} \, A$.Substituting these values into the equation:$10 - (40 	imes 10^{-6} 	imes 245 	imes 10^3) - V_{BE} = 0$$10 - (40 	imes 245 	imes 10^{-3}) - V_{BE} = 0$$10 - (9800 	imes 10^{-3}) - V_{BE} = 0$$10 - 9.8 - V_{BE} = 0$$0.2 - V_{BE} = 0$$V_{BE} = 0.2 \, V$.

In the given common emitter transistor circuit,the base current is $I_B = 40 \, \mu A$. Determine the base-emitter voltage $V_{BE}$. (in $, V$)

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