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JEE Main 2023 · Physics · Semiconductors and Communication · easy

A zener diode of breakdown voltage 6 V is connected in parallel with a load of 600 Ω, and this combination is fed from a 12 V source through a series…

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A zener diode of breakdown voltage 6 V6\ \mathrm{V} is connected in parallel with a load of 600 Ω600\ \Omega, and this combination is fed from a 12 V12\ \mathrm{V} source through a series resistor of 200 Ω200\ \Omega. Assuming the diode operates in breakdown, the power dissipated in the zener diode is
  1. A. 0.12 W0.12\ \mathrm{W}
  2. B. 0.18 W0.18\ \mathrm{W}
  3. C. 0.06 W0.06\ \mathrm{W}
  4. D. 0.24 W0.24\ \mathrm{W}
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Answer: (A) 0.12 W0.12\ \mathrm{W}

Voltage across the series resistor =12−6=6 V=12-6=6\ \mathrm{V}, so the total current is 6200=30 mA\dfrac{6}{200}=30\ \mathrm{mA}. The load takes 6600=10 mA\dfrac{6}{600}=10\ \mathrm{mA}; the rest flows through the zener: IZ=20 mAI_Z=20\ \mathrm{mA}. Power in the zener =VZIZ=6×0.02=0.12 W=V_ZI_Z=6\times0.02=0.12\ \mathrm{W}. Using the full 30 mA30\ \mathrm{mA} gives 0.18 W0.18\ \mathrm{W}.
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