Giving 2V to LED passes a 10 mA current. If you want to connect this diode to a 6V battery, then the value of resistance in series kept……

Options

A. `400 Omega`
B. `4000Omega`
C. `40Omega`
D. `300Omega`

Answer

A

Explanation

`400Omega`
Suppose initially the resistance of LED is `R_(1)`
`therefore R_(1) = (V)/(I)=(2)/(10xx10^(-3))=200Omega`
Now, if there is resistance `R_(2)` by connecting the LED to the 6V battery,
In V = IR, I is constant
`therefore V prop R`
`therefore (V_(2))/(V_(1))=(R_(2))/(R_(1))`
`(6)/(2)=(R_(2))/(200)`
`therefore R_(2)=600Omega`
`therefore` Resistance joining in series
`=R_(2)-R_(1)`
`= 600-200=400Omega`