For the circuit shown below, the current through the Zener diode is - (a) 5 mA (b) zero (c) 14 mA (d) 9 mA

Answer: (d). Without the Zener the load gets 120 × 10/15 = 80 V > 50 V, so the Zener holds 50 V. I = (120 - 50)/(5 kΩ) = 14 mA IL = (50)/(10 kΩ) = 5 mA IZ = 14 - 5 = 9 mA, so the answer is (d).

For the circuit shown below, the current through the Zener diode is - | JEE Main 2019 (10 Jan, Shift 2)

Q84JEE Main 2019 · 10 JAN · SHIFT 2Medium

Special Purpose Diodes

For the circuit shown below, the current through the Zener diode is -

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Answer: (d)

Without the Zener the load gets $120 \times \frac{10}{15} = 80$ V $> 50$ V, so the Zener holds $50$ V.
$I = \frac{120 - 50}{5\text{ k}\Omega} = 14$ mA
$I_L = \frac{50}{10\text{ k}\Omega} = 5$ mA
$I_Z = 14 - 5 = 9$ mA, so the answer is (d).

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