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JEE Main 2021 (20 Jul, Shift 2) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2021 (20 Jul, Shift 2)

Q69JEE Main 2021 · 20 JUL · SHIFT 2Medium

Special Purpose Diodes

A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance Rp is .................... Ω\Omega.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 192

  1. Maximum Zener current: I=PVZ=0.58=62.5I = \frac{P}{V_Z} = \frac{0.5}{8} = 62.5 mA
  2. At the full potential divider output 2020 V, drop across RpR_p is 20−8=1220 - 8 = 12 V.
  3. Rp=120.0625=192 ΩR_p = \frac{12}{0.0625} = 192\ \Omega
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