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A diode made forward biased by a two vol...


A diode made forward biased by a two volt battery however there is a drop of 0.5 V across the diode which is independent of current. Also a current greater then 10 mA produces large joule loss and damages diode. If diode is to be operated at 5 mA, the series resistance to be put is

A

`3 k Omega`

B

`300 k Omega`

C

`300 Omega`

D

`200 k Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=(V-V_(b))/(R)rArr510^(-2)=((2=0.5))/(R)rArrR=300 Omega`
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