In the figure, given that \(V_{BB}\) supply can vary from \(0\) to \(5.0 \,V,\,\,V_{CC} = 5\,V,\) \(\beta _{dc} = 200,\,\,R_B = 100,\,k\Omega ,\) \(R_C = 1\,k\Omega  \) an \(V_{BE} = 1.0\,V,\) The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively

\(\mathrm{V}_{\mathrm{CE}}=0\)

\(\mathrm{V}_{\mathrm{CC}}-\mathrm{R}_{\mathrm{C}} \mathrm{I}_{\mathrm{C}}=0\)

\(\mathrm{i}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{C}}}\) \(=5 \times 10^{-3} \mathrm{A}\)

\(I_C\,=\,Bi_B\)

\(\Rightarrow  \mathrm{i}_{\mathrm{B}}=25 \mu \mathrm{A} \)

\( \Rightarrow \mathrm{V}_{\mathrm{BB}}=\mathrm{i}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}-\mathrm{V}_{\mathrm{BE}}=0\)

\(\Rightarrow V_{\mathrm{BE}}=\mathrm{V}_{\mathrm{BE}}+\mathrm{i}_{\mathrm{BE}}=3.5 \mathrm{V}\)