1.

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……

Answer»

`400 Omega`
`4000Omega`
`40Omega`
`300OMEGA`

SOLUTION :`400OMEGA`
Suppose initially the resistance of LEDis `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
`thereforeV 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`


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