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(B) Applying Kirchhoff's voltage law to the base-emitter loop:$V_{BB} = I_B R_B + V_{BE} + I_E R_E$Since there is no base resistor $R_B$ in the circui

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$5 	imes 10^{-4} \ mA$

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(B) Applying Kirchhoff's voltage law to the base-emitter loop:$V_{BB} = I_B R_B + V_{BE} + I_E R_E$Since there is no base resistor $R_B$ in the circuit diagram,the equation simplifies to:$V_{BB} = V_{BE} + I_E R_E$Given $V_{BB} = 1.5 \ V$,$V_{BE} = 0.7 \ V$,and $R_E = 17 \ k\Omega = 17 	imes 10^3 \ \Omega$:$1.5 = 0.7 + I_E (17 	imes 10^3)$$0.8 = I_E (17 	imes 10^3)$$I_E = \frac{0.8}{17 	imes 10^3} \ A = \frac{0.8}{17} \ mA \approx 0.047 \ mA = 47 	imes 10^{-3} \ mA$We know that $I_E = (\beta + 1) I_B$:$I_B = \frac{I_E}{\beta + 1} = \frac{0.047 \ mA}{101} \approx 0.000465 \ mA = 4.65 	imes 10^{-4} \ mA$Rounding to the nearest provided option,we get $I_B \approx 5 	imes 10^{-4} \ mA$.

For the given bipolar junction transistor,find the approximate value of $I_B$,if $V_{CC} = 9 \ V$,$V_{BB} = 1.5 \ V$,$R_C = 13 \ k\Omega$,$R_E = 17 \ k\Omega$,and $\beta = 100$. (Assume the transistor is in active mode and $V_{BE} = 0.7 \ V$)

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