Physics - Semiconductors Question with Solution | TestHub

${\text{I}}_{\text{L}}=\frac{{\text{V}}_{\text{AB}}}{{\text{R}}_{\text{L}}}=\frac{10}{2000}=5\text{mA}$

${\text{I}}_{}=\frac{60-10}{4000}=\frac{50}{4000}=12·5\text{mA}$

$\text{I}={\text{I}}_{\text{L}}+{\text{I}}_{\text{Z}}\therefore {\text{I}}_{\text{Z}}=\text{I}-{\text{I}}_{\text{L}}=7·5\text{mA}$

Hence, 12.5 mA, 7.5 mA, 5 mA