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(D) In the input circuit,applying Kirchhoff's Voltage Law $(KVL)$ to the base-emitter loop:$V_{BB} - I_b R_b - V_{BE} = 0$Assuming the transistor is i

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$I_b = 10 	ext{ } \mu	ext{A}$

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(D) In the input circuit,applying Kirchhoff's Voltage Law $(KVL)$ to the base-emitter loop:$V_{BB} - I_b R_b - V_{BE} = 0$Assuming the transistor is in the active region and $V_{BE} \approx 0.7 	ext{ V}$,we need $V_{BB}$ to find $I_b$. However,looking at the circuit,the base current is determined by the input bias. Given the standard approach for such problems where $V_{BB}$ is often assumed to be equal to $V_{CC}$ if not specified,or by calculating the required $I_b$ to reach saturation:For saturation,$V_{CE} = 0 	ext{ V}$.Applying $KVL$ to the output loop:$V_{CC} - I_c R_c - V_{CE} = 0$$1 	ext{ V} - I_c (1 	imes 10^3 	ext{ } \Omega) - 0 = 0$$I_c = \frac{1}{1000} 	ext{ A} = 1 	ext{ mA} = 1000 	ext{ } \mu	ext{A}$.Since $\beta = \frac{I_c}{I_b}$,the base current required for saturation is:$I_b = \frac{I_c}{\beta} = \frac{1000 	ext{ } \mu	ext{A}}{100} = 10 	ext{ } \mu	ext{A}$.

For a given transistor amplifier circuit in $CE$ configuration with $V_{CC} = 1 \text{ V}$,$R_c = 1 \text{ k}\Omega$,$R_b = 100 \text{ k}\Omega$,and $\beta = 100$,the value of the base current $I_b$ is:

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