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Question

Reverse breakdown voltage for Zener diode series is 3.2 V, Find the current (i) through Zener diode.
  1. 52 mA
  2. 40 mA
  3. 42 mA
  4. 36 mA


Solution

The correct option is A 52 mA
Current through 400 Ω resistor =3.2400=8 mA
Apply KVL on the left loop,
icell=143.2180=10.8180=60 mA
i=iZener=608=52 mA

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