Physics · Semiconductor and Electronic Devices

JEE Main 2026 — 4 April, Evening Shift — Question 20

A diode has Zener voltage of $10\ \mathrm{V}$ and maximum power dissipation of $0.5\ \mathrm{W}$ , then the minimum resistance to be used in series with this diode for safety when it is connected to a $25\ \mathrm{V}$ power supply is $\Omega$ .

Answer: 300

Numerical answer — enter this value.

Step-by-step solution

Maximum current $I = P/V_z = 0.5/10 = 0.05\ \mathrm{A}$ . Voltage across resistor $= 25-10=15\ \mathrm{V}$ . So $R = 15/0.05 = 300\ \Omega$ .

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Semiconductor and Electronic Devices
Topic: p-n Diode and its Applications

Assertion A: A diode under reverse-biased condition provides very small current which is nearly independent of voltage until a critical…

Answer: 300

Step-by-step solution

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