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The reverse breakdown voltage of a Zener...

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit then `I_z` in mA is ___________

A

17m A

B

15mA

C

10mA

D

7mA

Text Solution

Verified by Experts

The correct Answer is:
C

For Zener break down potential difference across `800Omega` resistor will be 5.6V
`V_z=5.6V`
`i_2 =V_z/800 = (5.6)/800 = 7mA`
`DeltaV` across `200Omega = 9-5.6 = 3.4 V `
`i_(1)=(3.4)/(200)=17mA`
`i_1=i_2 +i_z`
`i_z=17mA -7mA=10mA`
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