Q135 / 180 · AIIMS 2016 · Physics

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?

  • Correct answer
    A
    $40 \mathrm{~mA}$
  • B
    $60 \mathrm{~mA}$
  • C
    $50 \mathrm{~mA}$
  • D
    $45 \mathrm{~mA}$

Solution

The maximum permissible current is $I_{Z_{\max }}=\frac{P}{V_Z}=\frac{364 \times 10^{-3}}{9.1}=40 \mathrm{~mA}$

Hence, the correct answer is option: 'A'.

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