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Lenz's Law

Reverse break...

Question

Reverse breakdown voltage for Zener diode series is $3.2 V$ , Find the current $(i)$ through Zener diode.

52 m A

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40 m A

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42 m A

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36 m A

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Solution

The correct option is A $52 m A$
Current through $400 Ω$ resistor $= \frac{3.2}{400} = 8 m A$
Apply $K V L$ on the left loop,
$i_{cell} = \frac{14 - 3.2}{180} = \frac{10.8}{180} = 60 m A$
$i = i_{Z e n e r} = 60 - 8 = 52 mA$

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