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JEE Main 2019 (8 Apr, Shift 1) · Special Purpose Diodes

Special Purpose Diodes question from JEE Main 2019 (8 Apr, Shift 1)

Q82JEE Main 2019 · 8 APR · SHIFT 1Medium

Special Purpose Diodes

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.

The current IZ through the Zener is:

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Answer: (d)

$I = \frac{9 - 5.6}{200} = 17$ mA
$I_L = \frac{5.6}{800} = 7$ mA
$I_Z = 17 - 7 = 10$ mA
So the answer is (d) $10$ mA.

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