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

Question

Aicharya · Accepted Answer

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 $\begin{aligned} \mathrm{I}_{\mathrm{z}} & =\mathrm{I}_1-\mathrm{I}_2 \\ & =(30-5) \mathrm{mA} \quad \ldots[\text { From (i) and (ii) }] \\ & =25 \mathrm{~mA} \end{aligned}$

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

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