JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2024
- Hard
- MCQ
In the given circuit, the breakdown voltage of the Zener diode is \(3.0 \mathrm{~V}\). What is the value of \(\mathrm{I}_z\) ?

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Answer & Solution
Correct Answer
(B) \(5.5 \mathrm{~mA}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}_z=3 \mathrm{~V}\)
Let potential at \(B=0 \mathrm{~V}\)
Potential at \(E\left(V_E\right)=10 \mathrm{~V}\)
\( V_C=V_A=3 \mathrm{~V} \)
\( I_z+I_1=I \)
\( I=\frac{10-3}{1000}=\frac{7}{1000} \mathrm{~A} \)
\( I_1=\frac{3}{2000} \mathrm{~A}\)
Therefore \(I_z=\frac{7-1.5}{1000}=5.5 \mathrm{~mA}\)
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