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(B) From the given circuit diagram,the base-emitter loop consists of a voltage source $V_{CC} = 10 \ V$,a base resistor $R_B = 245 \ k\Omega$,and the

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$0.2$

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(B) From the given circuit diagram,the base-emitter loop consists of a voltage source $V_{CC} = 10 \ V$,a base resistor $R_B = 245 \ k\Omega$,and the base-emitter junction $V_{BE}$.Applying Kirchhoff's Voltage Law $(KVL)$ to the base loop:$V_{CC} - I_B R_B - V_{BE} = 0$Given $I_B = 40 \ \mu A = 40 	imes 10^{-6} \ A$ and $R_B = 245 \ k\Omega = 245 	imes 10^3 \ \Omega$.Substituting the values:$10 - (40 	imes 10^{-6} 	imes 245 	imes 10^3) - V_{BE} = 0$$10 - (40 	imes 0.245) - V_{BE} = 0$$10 - 9.8 - V_{BE} = 0$$0.2 - V_{BE} = 0$$V_{BE} = 0.2 \ V$.

In a common emitter transistor circuit,the base current is $40 \ \mu A$. Find the value of $V_{BE}$. (in $V$)

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