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A two volt battery forward biased and a ...

A two volt battery forward biased and a diode. However, there is a drop of 0.5 V across the diode which is independent of current. Also, a current greater than 10 mA produces large joule loss and damages diode. If diode is to be operated at 5 mA, then 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
By KVL, `2=0.5 + IR=0.5 + 5 xx 10^(-3)`R
`R=(15)/(5 xx 10^(-3)`, `R=1500/5=300Omega`
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