The reverse breakdown voltage of a Zener diode is $5.6 V$ in the given circuit.
The current $I_{Z}$ through the Zener is

7 m A

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17 m A

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10 m A

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15 m A

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Solution

The correct option is C $10 m A$
$9 = V_{Z} + V_{R_{1}}$
$V_{Z} = 5.6 V$
$V_{R_{1}} = 9 - 5.6$
$V_{R_{1}} = 3.4$
$I_{R_{1}} = \frac{V_{R_{1}}}{R} = \frac{3.4}{200}$
$I_{R_{1}} = 17 m A$
$V_{Z} = V_{R_{2}} = I_{R_{2}} (R_{2})$
$\frac{5.6}{800} = I_{R_{2}}$
$I_{R_{2}} = 7 m A$
$I_{Z} = (17 - 7) m A$
$= 10 m A$ .

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