In the figure shown, the maximum current through Zener diode is 9 mA, then the Zener breakdown voltage is

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20 V

30 V

40 V

50 V

answer is D.

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Detailed Solution

$I_{\max} = \frac{120 - V_{z}}{5 \times 10^{3}}$

$\begin{array}{l} {(I_{z})}_{\max} = I_{\max} - I_{L} \\ 9 m A = (\frac{120 - V_{z}}{5 \times 10^{3}}) - \frac{V_{z}}{10 \times 10^{- 3}} \\ \Rightarrow V_{z} = 50 V \end{array}$

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