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

The reverse break down voltage of a Zener diode is $5.6 \mathrm{V}$ in the given circuit.

The current ${\mathrm{I}}_z$ through the Zener is:

1. $10 \mathrm{m}\mathrm{A}$
2. $7 \mathrm{m}\mathrm{A}$
3. $17 \mathrm{m}\mathrm{A}$
4. $15 \mathrm{m}\mathrm{A}$

Solution

${\mathrm{i}}_1-{\mathrm{i}}_z=\frac{{\mathrm{V}}_{\mathrm{z}}}{800}=\frac{5.6}{800}=7\mathrm{ }\mathrm{m}\mathrm{A}$
For loop ABCF,
$9-200\mathrm{ }{\mathrm{i}}_1-{\mathrm{V}}_{\mathrm{z}}=0$
${\mathrm{i}}_1=0.017\mathrm{ }\mathrm{o}\mathrm{r}\mathrm{ }17\mathrm{ }\mathrm{m}\mathrm{A}$
${\mathrm{i}}_{\text{z}}=\left({\mathrm{i}}_1-7\right)\mathrm{ }\mathrm{m}\mathrm{A}=10\mathrm{ }\mathrm{m}\mathrm{A}$

Asked in: JEE Main 2019 (08 Apr Shift 1)

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