JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 20)
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 $$\Omega$$. In the given circuit RL represents the variable load resistance. The ratio of maximum and minimum value of RL is _____________.
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Answer
2
Explanation
Minimum value of RL for which the diode is shorted is $${{{R_L}} \over {{R_L} + 100}} \times 10 = 8 \Rightarrow {R_L} = 400\,\Omega $$
For maximum value of RL, current through diode is 10 mA
So $${i_R} = {i_{{R_L}}} + {I_{ZM}}$$
$${2 \over {100}} = {8 \over {{R_L}}} + 10 \times {10^{ - 3}}$$
$$10 \times {10^{ - 3}} = {8 \over {{R_L}}}$$
$${R_L} = 800\,\Omega $$
So $${{{R_{L\,\max }}} \over {{R_{L\,\min }}}} = 2$$
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