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JEE Main 2021 · Physics · Semiconductors and Communication · easy

The input is 15 V, the series resistor is 500 Ω, and a load of 1.5 kΩ is connected across the Zener.

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A Zener diode of breakdown voltage 9 V9\ \mathrm{V} is used as a voltage regulator. The input is 15 V15\ \mathrm{V}, the series resistor is 500 Ω500\ \Omega, and a load of 1.5 kΩ1.5\ \mathrm{k\Omega} is connected across the Zener. The current through the Zener diode is
  1. A. 12 mA12\ \mathrm{mA}
  2. B. 18 mA18\ \mathrm{mA}
  3. C. 9 mA9\ \mathrm{mA}
  4. D. 6 mA6\ \mathrm{mA}
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Answer: (D) 6 mA6\ \mathrm{mA}

Voltage across the series resistor =15−9=6 V=15-9=6\ \mathrm{V}, so the total current is I=6500=12 mAI=\dfrac{6}{500}=12\ \mathrm{mA}. Load current =91500=6 mA=\dfrac{9}{1500}=6\ \mathrm{mA}. The Zener carries the rest: 12−6=6 mA12-6=6\ \mathrm{mA}.
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