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QJEEM22Phe09b: JEE Main 2022 Physics - Class 12 JEEM22Phe09b Solution: A zener of breakdown voltage V z = 8 V and maximum zener…

A zener of breakdown voltage Vz = 8 V and maximum zener current, IZM = 10 mA is subjectd to an input voltage Vi = 10 V with series resistance R = 100 Ω\Omega . In the given circuit RL represents the variable load resistance. The ratio of maximum and minimum value of RL is _____________.

JEE Main 2022 (Online) 28th June Evening Shift Physics - Semiconductor Question 84 English

Step-by-Step Verified Solution

Minimum value of RL for which the diode is shorted is RLRL+100×10=8⇒RL=400 Ω{{{R_L}} \over {{R_L} + 100}} \times 10 = 8 \Rightarrow {R_L} = 400\,\Omega

For maximum value of RL, current through diode is 10 mA

So iR=iRL+IZM{i_R} = {i_{{R_L}}} + {I_{ZM}}

2100=8RL+10×10−3{2 \over {100}} = {8 \over {{R_L}}} + 10 \times {10^{ - 3}}

10×10−3=8RL10 \times {10^{ - 3}} = {8 \over {{R_L}}}

RL=800 Ω{R_L} = 800\,\Omega

So RL max⁡RL min⁡=2{{{R_{L\,\max }}} \over {{R_{L\,\min }}}} = 2

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