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The reverse breakdown voltage of a Zener...

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

The current `I_(2)` through the Zener is:

Text Solution

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The correct Answer is:
10
Assuming no diode potential drop across `800Omega` resistor = 7.2 V
`implies` Zener breakdown has occurred `implies` Current through `800Ommega` resistor `=(5.6)/(800)A`
`implies` Current through `200Omega` resistor `=(9-5.6)/(200)=3.4/200`
`implies` Current through Zener diode `=(3.4)/(200)-5.6/800=8/800implies10mA`
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