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JEE Main 2026 (23 Jan, Shift 1) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2026 (23 Jan, Shift 1)

Q12JEE Main 2026 · 23 JAN · SHIFT 1Medium

Special Purpose Diodes

The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at Vz=5 VV_z = 5\text{ V} and the desired current in load is 5 mA. The unregulated voltage source can supply upto 25 V. Considering the Zener diode can withstand four times of the load current, the value of resistor RsR_s (shown in circuit) should be ‾ Ω\underline{\qquad\quad}\ \Omega.

Figure

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Answer: (c)

  1. Worst case is the maximum input 2525 V, giving drop 25−5=2025 - 5 = 20 V across RsR_s.
  2. Safe current limit: 4IL=4×5=204 I_L = 4 \times 5 = 20 mA, taken as the current through RsR_s.
  3. Rs=2020×10−3=1000 ΩR_s = \frac{20}{20 \times 10^{-3}} = 1000\ \Omega
  4. So the answer is (c) 1000 Ω1000\ \Omega.
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