For a given transistor amplifier circuit in CE configuration, ${\mathrm{V}}_{\mathrm{CC}}=1 \mathrm{V}$, ${\mathrm{R}}_{\mathrm{C}}=1 \mathrm{k\Omega }$, ${\mathrm{R}}_{\mathrm{b}}=100 \mathrm{k\Omega }$ and $\mathrm{\beta }=100$ . Value of base current I_b is

Given that
$\begin{array}{l}{\mathrm{v}}_{\mathrm{cc}}=1\mathrm{V},{\mathrm{R}}_{\mathrm{c}}=1\mathrm{k\Omega }, {\mathrm{R}}_{\mathrm{b}}=100\mathrm{k\Omega }, \mathrm{\beta }=100\\ \Rightarrow \frac{{\mathrm{I}}_{\mathrm{c}}}{{\mathrm{I}}_{\mathrm{b}}}=100\end{array}$

By using RVL rule,
${\mathrm{V}}_{\mathrm{CC}}-{\mathrm{i}}_{\mathrm{C}}{\mathrm{R}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{CE}}=0$
But saturation conditions, ${\mathrm{V}}_{\mathrm{CE}}\to 0$, 

$\Rightarrow 1={\mathrm{i}}_{\mathrm{c}}\times {10}^3\Rightarrow {\mathrm{i}}_{\mathrm{c}}=\frac{1}{{10}^3} \mathrm{A}$
$\begin{array}{l}{\mathrm{I}}_{\mathrm{b}}=\frac{{\mathrm{I}}_{\mathrm{c}}}{100}=\frac{1}{{10}^3\times 100}={10}^{-5} \mathrm{A}\\ \Rightarrow {\mathrm{I}}_{\mathrm{b}}=10\times {10}^{-6} \mathrm{A}=10\mathrm{\mu A}\end{array}$