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MHT CET 2026 (19 May, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from MHT CET 2026 (19 May, Shift 2)

Q1MHT CET 2026 · 19 MAY · SHIFT 2Easy

Special Purpose Diodes

To use Zener diode of maximum power rating 1.01.0 W as a 4.04.0 V stabilized power supply, the maximum current flowing through Zener diode will

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Answer: (c)

  1. Zener power P=VZImaxP = V_Z I_{max}, so Imax=PVZI_{max} = \dfrac{P}{V_Z}.
  2. Imax=1.04.0=0.25 A=250 mAI_{max} = \dfrac{1.0}{4.0} = 0.25\ \text{A} = 250\ \text{mA}
  3. So the answer is (c) 250250 mA.
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