In the following circuit, the current flowing through zener diode is

The voltage drop across \(R_2\) is
\(\mathrm{V}_2=\mathrm{V}_{\mathrm{z}}=8 \mathrm{~V}\)
The current through \(R_2\) is
\(\begin{aligned}
& \mathrm{I}_2=\frac{\mathrm{V}_2}{\mathrm{R}_2}=\frac{8}{1600}=5 \times 10^{-3} \mathrm{~A} \\
& \mathrm{I}_2=5 \mathrm{~mA}
\end{aligned}...(i)\)
The voltage drop across \(R_1\) is
\(\begin{aligned}
& \mathrm{V}_1=20-\mathrm{V}_2 \\
& 20-8=12 \mathrm{~V}
\end{aligned}\)
The current through \(R_1\) is
\(\mathrm{I}_1=\frac{\mathrm{V}_1}{\mathrm{R}_1}=\frac{12}{400}=3 \times 10^{-2} \mathrm{~A}=30 \times 10^{-3} \mathrm{~A}\)
\(\mathrm{I}_1=30 \mathrm{~mA}...(ii)\)
The current through the Zener diode is
\(\begin{aligned}
\mathrm{I}_{\mathrm{z}} & =\mathrm{I}_1-\mathrm{I}_2 \\
& =(30-5) \mathrm{mA} \quad \ldots[\text { From (i) and (ii) }] \\
& =25 \mathrm{~mA}
\end{aligned}\)