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(B) Given: $\beta = 50$,$I_B = 40 \, \mu A = 40 	imes 10^{-6} \, A$,$V_{CC} = 10 \, V$,$R_C = 2 \, k\Omega = 2000 \, \Omega$.Step $1$: Calculate the

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$6$

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(B) Given: $\beta = 50$,$I_B = 40 \, \mu A = 40 	imes 10^{-6} \, A$,$V_{CC} = 10 \, V$,$R_C = 2 \, k\Omega = 2000 \, \Omega$.Step $1$: Calculate the collector current $I_C$.$I_C = \beta 	imes I_B = 50 	imes 40 	imes 10^{-6} \, A = 2000 	imes 10^{-6} \, A = 2 	imes 10^{-3} \, A = 2 \, mA$.Step $2$: Apply Kirchhoff's Voltage Law to the output loop.$V_{CE} = V_{CC} - I_C R_C$.Step $3$: Substitute the values.$V_{CE} = 10 \, V - (2 	imes 10^{-3} \, A) 	imes (2000 \, \Omega) = 10 \, V - 4 \, V = 6 \, V$.Therefore,the collector-emitter voltage $V_{CE}$ is $6 \, V$.

In the following transistor amplifier circuit,$\beta = 50$. The $V_{CE}$ of the transistor is .......... $V$.

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