The value of I_c for the configuration shown | vedclass

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Question

(A) Given:$V_{CC} = 12 \, V$$R_B = 240 \, k\Omega$$\beta = 75$$V_{BE} = 0.7 \, V$Step $1$: Calculate the base current $I_B$ using the input loop equat

Vedclass · Accepted Answer

$3.53$

Vedclass · Answer

(A) Given:$V_{CC} = 12 \, V$$R_B = 240 \, k\Omega$$\beta = 75$$V_{BE} = 0.7 \, V$Step $1$: Calculate the base current $I_B$ using the input loop equation:$I_B = \frac{V_{CC} - V_{BE}}{R_B} = \frac{12 \, V - 0.7 \, V}{240 	imes 10^3 \, \Omega} = \frac{11.3 \, V}{240 	imes 10^3 \, \Omega} = 47.08 	imes 10^{-6} \, A = 47.08 \, \mu A$Step $2$: Calculate the collector current $I_C$ using the relation $I_C = \beta I_B$:$I_C = 75 	imes 47.08 \, \mu A = 3531 \, \mu A = 3.53 \, mA$Thus,the value of $I_C$ is $3.53 \, mA$.

The value of $I_c$ for the configuration shown is in $mA$.

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