JEE Main 2026 Apr 4 Shift 2, Physics Q21: Semiconductors, p-n Junction Diode and Applications

JEE Main2026Apr 4, Shift 2Physics

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

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Solution

Maximum permissible current through the Zener: $i_{ma x} = \frac{P}{V _{Z}} = \frac{0.5}{10} = 0.05 A$ .

The series resistor drops the difference between supply and Zener voltages at this current:

$V_{s u ppl y} = i_{ma x} R + V_{Z} \Rightarrow 25 = 0.05 R + 10$ .

$R_{min} = \frac{15}{0.05} = 300 Ω$ .

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JEE Main 2026 Apr 4 Shift 2, Physics Q21: Semiconductors, p-n Junction Diode and Applications

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