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JEE Main 2021 (20 Jul, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2021 (20 Jul, Shift 2)

Q69JEE Main 2021 · 20 JUL · SHIFT 2Medium

Special Purpose Diodes

A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance Rp is .................... $\Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 192

Maximum Zener current: $I = \frac{P}{V_Z} = \frac{0.5}{8} = 62.5$ mA
At the full potential divider output $20$ V, drop across $R_p$ is $20 - 8 = 12$ V.
$R_p = \frac{12}{0.0625} = 192\ \Omega$

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