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In the figure shown, the currents through the zener diode and load resistance are respectively

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2mA, 6 mA

4 mA, 2 mA

6 mA, 4 mA

2 mA, 8 mA

answer is B.

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Detailed Solution

$I = \frac{90}{15 k} = 6 m A$
$I_{L} = \frac{10}{5 K} = 2 m A$
$I_{Z} = I - I_{2} = 4 m A$

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