JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 18)

In the following circuit, the reading of the ammeter will be

$($ Take Zener breakdown voltage $=4 \mathrm{~V})$

60 mA

80 mA

10 mA

24 mA

JEE Main 2025 (Online) 7th April Morning Shift Physics - Semiconductor Question 6 English Explanation

$$\mathrm{V}_1=\frac{400}{100+400} \times 12 \mathrm{~V}=\frac{4}{5} \times 12=\frac{48}{5} \mathrm{~V}$$

here, $\mathrm{V}_1>\mathrm{V}_{\mathrm{z}},\left(\mathrm{V}_{\mathrm{z}}=\right.$ Zener Voltage $)$

So, Zener breakdown will be take place So, voltage across $400 \Omega$ will be 4 V

$$\mathrm{I}=\frac{4}{400} \mathrm{~A}=\frac{1}{100 \mathrm{~A}}=10 \mathrm{~mA}$$