JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2019
- Hard
- Numerical
The reverse breakdown voltage of a Zener diode is \(5.6\, V\) in the given circuit. The current \(I_z\) through the Zener is......\(mA\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(9=V_{z}+V_{R_{1}}\)
\(V_{z}=5.6 v\)
\(V_{R_1}=9-5.6\)
\(V_{R_1}=3.4\)
\(\mathrm{I}_{\mathrm{R}_{1}}=\frac{\mathrm{V}_{\mathrm{R}_{1}}}{\mathrm{R}}=\frac{3.4}{200} ; \mathrm{I}_{\mathrm{R}_{1}}=17 \mathrm{mA}\)
\({{\text{V}}_{\text{z}}}={{\text{V}}_{{{\text{R}}_{2}}}}={{\text{I}}_{{{\text{R}}_{2}}}}\left( {{\text{R}}_{2}} \right)\)
\(\frac{5.6}{800}={{\text{l}}_{{{\text{R}}_{2}}}}\) ; \({{I}_{{{\text{R}}_{2}}}}=7\text{mA}\)
\({{I}_{\text{z}}}=(17-7)\,\text{mA}=10\,\text{mA}\)
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