b
The voltage drop across \(\mathrm{R}_{2}\) is \(\mathrm{V}_{\mathrm{R}_{2}}\) \(=\mathrm{V}_{\mathrm{Z}}=10 \mathrm{V}\)

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

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