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

Options

A. `3 k Omega`
B. `300 k Omega`
C. `300 Omega`
D. `200 k Omega`

Answer

C

Explanation

`I=(V-V_(b))/(R)rArr510^(-2)=((2=0.5))/(R)rArrR=300 Omega`