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MHT CET 2024 (10 May, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from MHT CET 2024 (10 May, Shift 2)

Q30MHT CET 2024 · 10 MAY · SHIFT 2Medium

Special Purpose Diodes

In the given circuit, Zener breakdown voltage is 8 V. If power of Zener diode is 1.6 W. The value of RR is

Figure

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Answer: (d)

  1. Maximum Zener current: IZ=PVZ=1.68=0.2I_Z = \frac{P}{V_Z} = \frac{1.6}{8} = 0.2 A
  2. Drop across RR: 10−8=210 - 8 = 2 V (no load, so all of it flows through the Zener).
  3. R=20.2=10 ΩR = \frac{2}{0.2} = 10\ \Omega
  4. So the answer is (d) 10 Ω10\ \Omega.
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