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(A) Given:$\beta = 48$$I_{B} = 200 \mu A = 200 	imes 10^{-6} \ A$$R_{C} = 500 \ \Omega$$V_{CC} = 5 \ V$Step $1$: Calculate the collector current $I_{

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$0.2$

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(A) Given:$\beta = 48$$I_{B} = 200 \mu A = 200 	imes 10^{-6} \ A$$R_{C} = 500 \ \Omega$$V_{CC} = 5 \ V$Step $1$: Calculate the collector current $I_{C}$.$I_{C} = \beta I_{B} = 48 	imes 200 	imes 10^{-6} \ A = 9600 	imes 10^{-6} \ A = 9.6 	imes 10^{-3} \ A = 9.6 \ mA$.Step $2$: Apply Kirchhoff's voltage law to the output loop.$V_{CC} = I_{C} R_{C} + V_{CE}$$V_{Y} = V_{CE} = V_{CC} - I_{C} R_{C}$$V_{Y} = 5 \ V - (9.6 	imes 10^{-3} \ A) 	imes (500 \ \Omega)$$V_{Y} = 5 \ V - 4.8 \ V = 0.2 \ V$.Thus,the voltage at terminal $Y$ is $0.2 \ V$.

In the circuit shown,the value of $\beta$ of the transistor is $48$. If the base current supplied is $200 \mu A$,what is the voltage at the terminal $Y$ (in $V$)?

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