The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit.

The current I_{z} through the Zener is :

 

 

  • Option 1)

    7 mA

  • Option 2)

    10 mA

  • Option 3)

    17 mA

  • Option 4)

    15 mA

Answers (1)

V_{Z}=5.6 V

So (I-I_{Z})=\frac{V_{Z}}{800}=\frac{5.6}{800}=7mA

So,(\Delta V)_{200\Omega }=9-5.6=3.4

& I = \frac{(\Delta V)_{200\Omega }}{200}=\frac{3.4}{200}=17 mA

So, I_{z} = 17-7=10 mA

I_{z} =10\; mA


Option 1)

7 mA

Option 2)

10 mA

Option 3)

17 mA

Option 4)

15 mA

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