Search any question & find its solution
Question:
Answered & Verified by Expert
A zener diode is specified as having a breakdown voltage of $9.1 \mathrm{~V}$, with a maximum power dissipation of $364 \mathrm{~mW}$. What is the maximum current the diode can handle?
Options:
Solution:
2693 Upvotes
Verified Answer
The correct answer is:
$40 \mathrm{~mA}$
The maximum permissible current is
$I_{Z_{\max }}=\frac{P}{V_Z}=\frac{364 \times 10^{-3}}{9.1}=40 \mathrm{~mA}$
$I_{Z_{\max }}=\frac{P}{V_Z}=\frac{364 \times 10^{-3}}{9.1}=40 \mathrm{~mA}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.