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JEE Main 2025 (2 Apr, Shift 1) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2025 (2 Apr, Shift 1)

Q22JEE Main 2025 · 2 APR · SHIFT 1Medium

Special Purpose Diodes

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 Ω400\,\Omega resistor connected in series, the zener current is found to be 4 times load current. The load current (IL)(I_L) and load resistance (RL)(R_L) are:

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Answer: (c)

  1. Current through series resistor: I=25−5400=50I = \frac{25 - 5}{400} = 50 mA
  2. I=IZ+IL=4IL+IL=5IL⇒IL=10I = I_Z + I_L = 4I_L + I_L = 5I_L \Rightarrow I_L = 10 mA
  3. RL=510×10−3=500 ΩR_L = \frac{5}{10 \times 10^{-3}} = 500\ \Omega
  4. So the answer is (c) IL=10I_L = 10 mA, RL=500 ΩR_L = 500\ \Omega.
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