1.

A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA when it operates with a 6 V battery through a limiting resistor R. The value of R is

Answer»

40 `k Omega`
`4 k Omega`
`200Omega`
`400 Omega`

Solution :CURRENT I = 10 mA = 0.01 A = `(6 - 2)/(R)rArr"" R = 400 Omega`


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