In the transistor circuit shown,the base curr | vedclass

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(B) From the given circuit diagram,the base-emitter loop consists of the $9 	ext{ V}$ battery and the base resistor $R_b$.Applying Kirchhoff's voltag

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$257$

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(B) From the given circuit diagram,the base-emitter loop consists of the $9 	ext{ V}$ battery and the base resistor $R_b$.Applying Kirchhoff's voltage law to the base-emitter loop (assuming the base-emitter voltage $V_{BE} \approx 0 	ext{ V}$ for an ideal transistor or that the $9 	ext{ V}$ is applied directly across $R_b$):$V_b = i_b 	imes R_b$Given $V_b = 9 	ext{ V}$ and $i_b = 35 \mu A = 35 	imes 10^{-6} 	ext{ A}$.$R_b = \frac{V_b}{i_b} = \frac{9}{35 	imes 10^{-6}} \Omega$$R_b = \frac{9}{35} 	imes 10^6 \Omega \approx 0.25714 	imes 10^6 \Omega$$R_b \approx 257.14 	imes 10^3 \Omega = 257.14 	ext{ k}\Omega$.Rounding to the nearest integer,we get $257 	ext{ k}\Omega$.

In the transistor circuit shown,the base current is $35 \mu A$. The value of the resistor $R_b$ is......$ k \Omega$.

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