Question

A light emitting diode has a voltage drop of $2V$ across it and passes a current of $1mA.$ If it operates with a $6V$ battery through a resistor $R.$ then value of $R$ is

• A. $400\Omega$
• B. $4k\Omega$
• C. $40k\Omega$
• D. $2k\Omega$

Answer 1

The correct option is A $400\Omega$

Light emitting diode is a forward biased $P-N$ junction which emits light$.$

The voltage across it $=2v$

Battery voltage $=6v$

Hence total voltage in the circuit $=V=6-2=4v$

since a LED has 0 resistance in the forward biased region, total resistance in the circuit$,$

$=0+r=r$

current $I=10mA=10/1000A=0.01A$

we know that$,$

$R=V/I$

$=>r=4/0.01=400\Omega$

So$,$ the value of the limiting resistor is $400\Omega$

Hence,

option $\left(A\right)$ is correct answer.