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In the figure, the reverse breakdown voltage of a Zener diode is 5.6 V, then the current $I_{z}$ through the diode is

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17 mA

15 mA

10 mA

7 Ma

answer is C.

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Detailed Solution

$I_{2} = \frac{5.6}{800} = 7 m A$

$\begin{array}{l} I = (\frac{9 - 5.6}{200}) = 17 m A \\ I_{Z} = I - I_{L} = 10 m A \end{array}$

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