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(D) The maximum power dissipation $(P)$ of the Zener diode is given as $364 \;mW = 364 	imes 10^{-3} \;W$.The breakdown voltage $(V_Z)$ is given as $

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$40$

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(D) The maximum power dissipation $(P)$ of the Zener diode is given as $364 \;mW = 364 	imes 10^{-3} \;W$.The breakdown voltage $(V_Z)$ is given as $9.1 \;V$.The maximum current $(I_{Z_{\max}})$ that the diode can handle is calculated using the formula:$I_{Z_{\max}} = \frac{P}{V_Z}$Substituting the given values:$I_{Z_{\max}} = \frac{364 	imes 10^{-3} \;W}{9.1 \;V}$$I_{Z_{\max}} = 40 	imes 10^{-3} \;A$Converting to milliamperes $(mA)$:$I_{Z_{\max}} = 40 \;mA$.

$A$ 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 $(mA)$ the diode can handle?

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