An LED has a voltage drop of 2 V across it and a current of 10 mA flows when it operates with a 6 V battery through a limiting resistance R. The value of R is:

$400 \Omega$

The correct answer is Option (4) → $400 \Omega$

To calculate the value of the limiting resistance R,

$R=\frac{V_R}{I}$ [Ohm's law]

$V_R=V_{battery}-V_{LED}=6V-2V=4V$

$R=\frac{V_R}{I}=\frac{4}{0.01}$

$=400Ω$