14. Semicondutor electronics — Physics STD 12 Science — Question

The current through $\mathrm{R}_{2}$ is

$\mathrm{I}_{\mathrm{R}_{2}}=$ $\frac{\mathrm{V}_{\mathrm{R}_{2}}}{\mathrm{R}_{2}}$ $=\frac{10 \mathrm{V}}{1500 \Omega}$ $=0.667 \times 10^{-2} \mathrm{A}$

$=6.67 \times 10^{-3} \mathrm{A}=6.67 \mathrm{mA}$

The voltage drop across $R_{1}$ is

$\mathrm{V}_{\mathrm{R}_{1}}=15 \mathrm{V}-\mathrm{V}_{\mathrm{R}_{2}}$ $=15 \mathrm{V}-10 \mathrm{V}=5 \mathrm{V}$

The current through $\mathrm{R}_{1}$ is

$\mathrm{I}_{\mathrm{R}_{1}}=\frac{\mathrm{V}_{\mathrm{R}_{1}}}{\mathrm{R}_{1}}=\frac{5 \mathrm{V}}{500 \Omega}=10^{-2} \mathrm{A}=10 \times 10^{-3} \mathrm{A}=10 \mathrm{mA}$

The current through the zener diode is

$\mathrm{I}_{\mathrm{Z}}=\mathrm{I}_{\mathrm{R}_{1}}-\mathrm{I}_{\mathrm{R}_{2}}$ $=(10-6.67) \mathrm{mA}=3.3 \mathrm{mA}$