Semiconductors

JEE Main level, single correct, medium

90 of 209

Question 90 of 209

The reverse break down voltage of a Zener diode is 5.6V5.6V in the given circuit.

The current
Iz{I}_{z} through the Zener is:

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Answer: (A) 10mA10mA


i1−iz=Vz800=5.6800=7mA{i}_{1}-{i}_{z}=\dfrac{{V}_{z}}{800}=\dfrac{5.6}{800}=7mA
For loop ABCF,

9−200i1−Vz=09-200{i}_{1}-{V}_{z}=0
i1=0.017or17mA{i}_{1}=0.017or17mA
iz=(i1−7)mA=10mA{i}_{\text{z}}=\left({i}_{1}-7\right)mA=10mA

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