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Question

An experiment is performed to determine the IV characteristics of a Zener diode, which has a protective resistance of R=100Ω, and a maximum power of dissipation rating of 1W. The minimum voltage range of the DC source in the circuit is

A
024V
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B
05V
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C
012V
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D
08V
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Solution

The correct option is A 024V
The circuit diode for a zener diode with a protective resistance is shown.

Applying KVL,
V=iR+VD
VD=ViR=V100i

Maximum power dissipated across the zener diode is given by:
P=VDi=(V100i)i=1
100i2Vi+10
Current should be real, hence determinant is greater than zero.
V2>400
V>20V

555925_474547_ans.gif

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