JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2022
- Hard
- Numerical
A Zener of breakdown voltage \(V_{z}=8\,V\) and maximum zener current, \(I _{ ZM }=10\,mA\) is subjected to an input voltage \(V _{ i }=10\,V\) with series resistance \(R=100\,\Omega\). In the given circuit \(R_{L}\) represents the variable load resistance. The ratio of maximum and minimum value of \(R_{L}\) is

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(I =\frac{2}{100}=20 mA\)
\(V _{ L }= I _{ L } R _{ L }\)
\(8=10 \times 10^{-3} \times R _{ L _{\max }}\)
\(\frac{4}{5} \times 10^{3}= R _{ L _{\max }}\)
\(800= R _{ L _{\max }}\)
\(I = I _{ Z }+ I _{ L }\)
\(I _{ L }=10 mA\)
If \(I _{ Z }=0\)
\(I _{ L _{\max }}=20 mA\)
\(V _{ L }= I _{ L _{\max }} \times R _{ L _{\min }}\)
\(\frac{8}{20} \times 10^{3}= R _{ L _{\min }}\)
\(400= R _{ L _{\min }}\)
\(\frac{ R _{ L _{\max }}}{ R _{ L _{\min }}}=\frac{800}{400}=2\)
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