The reverse break down voltage of a Zener diode is 5.6 V in the given circuit. The current I z through the Zener is:

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The reverse break down voltage of a Zener diode is

5.6 V

in the given circuit.

The current

I_{z}

through the Zener is:

PhysicsSemiconductorsJEE MainJEE Main 2019 (08 Apr Shift 1)

Options:

A $10 m A$
B $7 m A$
C $17 m A$
D $15 m A$

Solution:

2853 Upvotes Verified Answer

The correct answer is:

10 m A

i_{1} - i_{z} = \frac{V_{z}}{800} = \frac{5.6}{800} = 7 m A

For loop ABCF,

9 - 200 i_{1} - V_{z} = 0

i_{1} = 0.017 o r 17 m A

i_{z} = (i_{1} - 7) m A = 10 m A

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