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JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Physics
Chapter: Semiconductors

JEE Main 2025 · 7 April · Morning Shift · Physics · Semiconductors

In the following circuit, the reading of the ammeter will be (Take...

In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage

=4 \mathrm{~V})

In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage =4 ~V)

Options

A24 mA
B80 mA
C10 mA
D60 mA

Correct answer

C. 10 mA

Step-by-step solution

In the following circuit, the reading of the ammeter will be (Take Zener breakdown voltage =4 ~V)

\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}

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