In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage $=4 \mathrm{~V})$
- 24 mA
- 80 mA
- 10 mA
- 60 mA
Solution
$\mathrm{V}_1=\frac{400}{100+400} \times 12 \mathrm{~V}=\frac{4}{5} \times 12=\frac{48}{5} \mathrm{~V}$
here, $\mathrm{V}_1 \gt \mathrm{V}_{\mathrm{z}},\left(\mathrm{V}_{\mathrm{z}}=\right.$ Zener Voltage $)$ So, Zener breakdown will be take place So, voltage across $400 \Omega$ will be 4 V
$\mathrm{I}=\frac{4}{400} \mathrm{~A}=\frac{1}{100 \mathrm{~A}}=10 \mathrm{~mA}$
Asked in: JEE Main 2025 (07 Apr Shift 1)