In the circuit shown in the figure, the transistor used has a current gain $\beta$ =100 . When the bias resistance ${\mathrm{R}}_{\mathrm{B}} = \mathrm{a} \mathrm{x} {10}^5\mathrm{\Omega }, {\mathrm{V}}_{\mathrm{CE}} = 5\mathrm{V}$ . Neglecting ${\mathrm{V}}_{\mathrm{BE}}$ , find the value of a.

We know that

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ce}} = {\mathrm{V}}_{\mathrm{c}} - {\mathrm{I}}_{\mathrm{c}}{\mathrm{R}}_{\mathrm{c}} \\ 5 = 10- {\mathrm{I}}_{\mathrm{c}}\mathrm{x}{10}^3 \\ {\mathrm{I}}_{\mathrm{C}} \mathrm{x} {10}^3 = 5 \\ {\mathrm{I}}_{\mathrm{c}} = \frac{1}{200} \end{gathered}$$

And

$$\begin{gathered} \mathrm{\beta } = \frac{{\mathrm{I}}_{\mathrm{C}}}{{\mathrm{I}}_{\mathrm{B}}} \\ 100 =\frac{1}{200} \mathrm{x} \frac{1}{{\mathrm{I}}_{\mathrm{B}}} \\ {\mathrm{I}}_{\mathrm{B}} = \frac{1}{20000} \end{gathered}$$

Therefore, we have

${\mathrm{V}}_{\mathrm{BE}} = {\mathrm{V}}_{\mathrm{C}} - {\mathrm{I}}_{\mathrm{B}} \mathrm{x} {\mathrm{R}}_{\mathrm{B}}$

Neglecting ${\mathrm{V}}_{\mathrm{BE}}$ in the above question

$$\begin{gathered} {\mathrm{V}}_{\mathrm{C}} = {\mathrm{I}}_{\mathrm{B}}{\mathrm{R}}_{\mathrm{B}} \\ 10 = \frac{1}{20000} \mathrm{x} {\mathrm{R}}_{\mathrm{B}} \\ {\mathrm{R}}_{\mathrm{B}} = \mathrm{a} \mathrm{x} {10}^5 \\ 10 = \frac{1}{20000} \mathrm{x} \mathrm{a} \mathrm{x} {10}^5 \\ \mathrm{a} = \frac{200000}{100000} \\ \mathrm{a} = 2 \end{gathered}$$