JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2025
- Medium
- MCQ
In the following circuit, the reading of the ammeter will be : (Take Zener breakdown voltage \(=4 \mathrm{~V})\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 10 mA
Step-by-step Solution
Detailed explanation
\(\mathrm{V}_1=\frac{400}{100+400} \times 12 \mathrm{~V}=\frac{4}{5} \times 12=\frac{48}{5} \mathrm{~V}\)
here, \(\mathrm{V}_1 \gt \mathrm{V}_{\mathrm{z}},\left(\mathrm{V}_{\mathrm{z}}=\right.\) 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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