Question

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 6 V battery through a limiting resistor R. The value of R is

• A. 40 k$\Omega$
• B. 4 k$\Omega$
• C. 200 $\Omega$
• D. 400 $\Omega$

Answer 1

The correct option is D 400 $\Omega$

As the LED is connected in series with the limiting resistor R, the potential difference across R
= Battery voltage - voltage drop across LED
= 6 - 2 = 4 V
$\therefore R=\frac{V}{I}=\frac{4V}{10mA}=\frac{4V}{10\times {10}^{-3}A}=400\Omega$