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Q.
A Zener diode is specified as having a breakdown voltage of $9.1 \,V$, with a maximum power dissipation of $364\, mW$. What is the maximum current the diode can handle?
AIIMSAIIMS 2016Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
The maximum permissible current is
$I_{Z_{max}} = \frac{P}{V_{Z}}$
$= \frac{364\times 10^{-3}}{9.1}$
$ = 40\,mA$