In a common emitter transistor circuit,the ba | vedclass

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(B) Applying Kirchhoff's voltage law to the base-emitter loop:$V_{CC} - I_B R_B = V_{BE}$Given:$V_{CC} = 10 \, V$$I_B = 40 \,\mu A = 40 	imes 10^{-6}

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$0.2$

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(B) Applying Kirchhoff's voltage law to the base-emitter loop:$V_{CC} - I_B R_B = V_{BE}$Given:$V_{CC} = 10 \, V$$I_B = 40 \,\mu A = 40 	imes 10^{-6} \, A$$R_B = 245 \, k\Omega = 245 	imes 10^3 \, \Omega$Substituting the values:$V_{BE} = 10 - (40 	imes 10^{-6} 	imes 245 	imes 10^3)$$V_{BE} = 10 - (40 	imes 245 	imes 10^{-3})$$V_{BE} = 10 - (9800 	imes 10^{-3})$$V_{BE} = 10 - 9.8$$V_{BE} = 0.2 \, V$Thus,the correct option is $B$.

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

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