JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
The reverse breakdown voltage of a Zener diode is \(5.6\, V\) in the given circuit. The current \(I_z\) through the Zener is......\(mA\)
- A \(10\)
- B \(15\)
- C \(7\)
- D \(17\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(9=V_{z}+V_{R_{1}}\) \(V_{z}=5.6 v\) \(V_{R_1}=9-5.6\) \(V_{R_1}=3.4\) \(\mathrm{I}_{\mathrm{R}_{1}}=\frac{\mathrm{V}_{\mathrm{R}_{1}}}{\mathrm{R}}=\frac{3.4}{200} ; \mathrm{I}_{\mathrm{R}_{1}}=17 \mathrm{mA}\)…
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