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  • A zener of breakdown voltage  and maximum zener current,<img alt="\mathrm{

A zener of breakdown voltage \mathrm{V_{Z}=8 \mathrm{~V}} and maximum zener current,\mathrm{\mathrm{I}_{\mathrm{ZM}}=10 \mathrm{~mA}} is subjected to an input voltage \mathrm{\mathrm{V}_{i}=10 \mathrm{~V}} with series resistance \mathrm{R}=100 \; \Omega. In the given circuit \mathrm{\mathrm{R}_{\mathrm{L}}} represents the variable load resistance. The ratio of maximum and minimum value of \mathrm{\mathrm{R}_{\mathrm{L}}} is ___________.

 

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

6


Answers (1)

best_answer

The current through R

I=\frac{2}{100}=20 \mathrm{~mA}

Case I-  For R_{L} to max current through it should be minimum

So I_Z= 10 mA \ and \ \ I_L= 10 mA

\begin{aligned} &\mathrm{V}_{\mathrm{L}}=\mathrm{I}_{\mathrm{L}} \mathrm{R}_{\mathrm{L}} \\ &8=10 \times 10^{-3} \times \mathrm{R}_{\mathrm{L}_{\max }} \\ &\frac{4}{5} \times 10^{3}=\mathrm{R}_{\mathrm{L}_{\max }} \\ &800=\mathrm{R}_{\mathrm{L}_{\max }} \end{aligned}

Case II-  For R_{L} to min current through it should be maximum 

So I_Z= 0 \ mA \ and \ \ I_L= 20 mA

\begin{aligned} &\mathrm{V}_{\mathrm{L}}=\mathrm{I}_{\mathrm{L}_{\max }} \times \mathrm{R}_{\mathrm{L}_{\min }} \\ &\frac{8}{20} \times 10^{3}=\mathrm{R}_{\mathrm{L}_{\min }} \\ &400=\mathrm{R}_{\mathrm{L}_{\min }} \end{aligned}

\frac{\mathrm{R}_{\mathrm{L}_{\max }}}{\mathrm{R}_{\mathrm{L}_{\min }}}=\frac{800}{400}=2

 

Posted by

Ajit Kumar Dubey

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