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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

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Answer & Solution

Correct Answer

(B) \(2\)

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\(I =\frac{2}{100}=20 mA\)

\(V _{ L }= I _{ L } R _{ L }\)

\(8=10 \times 10^{-3} \times R _{ L _{\max }}\)

\(\frac{4}{5} \times 10^{3}= R _{ L _{\max }}\)

\(800= R _{ L _{\max }}\)

\(I = I _{ Z }+ I _{ L }\)

\(I _{ L }=10 mA\)

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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