d
\({{\text{V}}_{{\text{breaddwon }}}} = 6{\text{V}},\) \({{\text{R}}_{\text{L}}} = 4\,{\text{k}}\Omega ,\) \({{\text{R}}_{\text{i}}} = 1{\text{k}}\Omega \)

\({{\text{i}}_{\text{L}}} = \frac{6}{4} \times {10^{ - 3}} = \) \(1.5 \times {10^{ - 3}} = 1.5{\text{mA}}\)

\(\mathrm{i}_{\mathrm{i}}=2 \times 10^{-3}\)

\({\text{i}} = {{\text{i}}_1} - {{\text{i}}_{\text{L}}} = 0.5{\text{mA}} - \) minimum current

\(\mathrm{i}_{\mathrm{i}}=10 \times 10^{-3}=10 \mathrm{mA}\)

\(\operatorname{i_{max}}=8.5 \mathrm{mA}\)