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JEE Main 20198 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Morning Shift
Subject: Physics
Chapter: Semiconductors

JEE Main 2019 · 8 April · Morning Shift · Physics · Semiconductors

The reverse break down voltage of a Zener diode is 5.6V in the...

The reverse break down voltage of a Zener diode is

5.6V

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

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

The current

{I}_{z}

through the Zener is:

Options

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

Correct answer

A. $10mA$

Step-by-step solution

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

{i}_{1} - {i}_{z} = \dfrac{{V}_{z}}{800} = \dfrac{5.6}{800} = 7mA

For loop ABCF,

9 - 200{i}_{1} - {V}_{z} = 0

{i}_{1} = 0.017or17mA

{i}_{\text{z}} = \left({i}_{1} - 7\right)mA = 10mA

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