Physics · 2019
JEE Main · 8 April 2019, Shift 1 · Q28
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current I_z through the Zener is:
The reverse breakdown voltage of a Zener diode is $\displaystyle 5.6$ V in the given circuit.
The current $\displaystyle \mathrm{I}_{\mathrm{z}}$ through the Zener is :
The current $\displaystyle \mathrm{I}_{\mathrm{z}}$ through the Zener is :Official answer
(2)
$\displaystyle 10 \text { mA }$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.